ehartford/alpaca1337-7b-4bit
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```cpp
#include <vector>
#include <unordered_map>
std::vector<int> twoSum(std::vector<int>& nums, int target) {
std::unordered_map<int, int> map;
for (int i = 0; i < nums.size(); i++) {
int complement = target - nums[i];
if (map.find(complement) != map.end()) {
return {map[complement], i};
}
map[nums[i]] = i;
}
return {};
}
```
Explanation:
The algorithm leverages a hash map (unordered_map in C++, HashMap in Java, dictionary in Python, and Map in JavaScript). It iterates through the given 'nums' array and calculates the complementary value (target - current value). If the complementary value is already in the hash map, it means that we found a solution, and we return those indices. If the complement is not in the hash map, we store the current element in the hash map with its index. If the algorithm doesn't find the solution, it returns an empty array or throws an exception (in Java).
This approach has a time complexity of O(n) and a space complexity of O(n) as well. | LeetCode 1: Two Sum
two-sum
Given an array of integers `nums` and an integer `target`, return _indices of the two numbers such that they add up to `target`_.
You may assume that each input would have **_exactly_ one solution**, and you may not use the _same_ element twice.
You can return the answer in any order.
**Example 1:**
**Input:** nums = \[2,7,11,15\], target = 9
**Output:** \[0,1\]
**Explanation:** Because nums\[0\] + nums\[1\] == 9, we return \[0, 1\].
**Example 2:**
**Input:** nums = \[3,2,4\], target = 6
**Output:** \[1,2\]
**Example 3:**
**Input:** nums = \[3,3\], target = 6
**Output:** \[0,1\]
**Constraints:**
* `2 <= nums.length <= 104`
* `-109 <= nums[i] <= 109`
* `-109 <= target <= 109`
* **Only one valid answer exists.**
**Follow-up:** Can you come up with an algorithm that is less than `O(n2)` time complexity? | Solve the following problem in c++, and explain your solution. |
```java
import java.util.HashMap;
import java.util.Map;
public int[] twoSum(int[] nums, int target) {
Map<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < nums.length; i++) {
int complement = target - nums[i];
if (map.containsKey(complement)) {
return new int[]{map.get(complement), i};
}
map.put(nums[i], i);
}
throw new IllegalArgumentException("No two sum solution");
}
```
Explanation:
The algorithm leverages a hash map (unordered_map in C++, HashMap in Java, dictionary in Python, and Map in JavaScript). It iterates through the given 'nums' array and calculates the complementary value (target - current value). If the complementary value is already in the hash map, it means that we found a solution, and we return those indices. If the complement is not in the hash map, we store the current element in the hash map with its index. If the algorithm doesn't find the solution, it returns an empty array or throws an exception (in Java).
This approach has a time complexity of O(n) and a space complexity of O(n) as well. | LeetCode 1: Two Sum
two-sum
Given an array of integers `nums` and an integer `target`, return _indices of the two numbers such that they add up to `target`_.
You may assume that each input would have **_exactly_ one solution**, and you may not use the _same_ element twice.
You can return the answer in any order.
**Example 1:**
**Input:** nums = \[2,7,11,15\], target = 9
**Output:** \[0,1\]
**Explanation:** Because nums\[0\] + nums\[1\] == 9, we return \[0, 1\].
**Example 2:**
**Input:** nums = \[3,2,4\], target = 6
**Output:** \[1,2\]
**Example 3:**
**Input:** nums = \[3,3\], target = 6
**Output:** \[0,1\]
**Constraints:**
* `2 <= nums.length <= 104`
* `-109 <= nums[i] <= 109`
* `-109 <= target <= 109`
* **Only one valid answer exists.**
**Follow-up:** Can you come up with an algorithm that is less than `O(n2)` time complexity? | Solve the following problem in java, and explain your solution. |
```javascript
function twoSum(nums, target) {
const map = new Map();
for (let i = 0; i < nums.length; i++) {
const complement = target - nums[i];
if (map.has(complement)) {
return [map.get(complement), i];
}
map.set(nums[i], i);
}
return [];
}
```
Explanation:
The algorithm leverages a hash map (unordered_map in C++, HashMap in Java, dictionary in Python, and Map in JavaScript). It iterates through the given 'nums' array and calculates the complementary value (target - current value). If the complementary value is already in the hash map, it means that we found a solution, and we return those indices. If the complement is not in the hash map, we store the current element in the hash map with its index. If the algorithm doesn't find the solution, it returns an empty array or throws an exception (in Java).
This approach has a time complexity of O(n) and a space complexity of O(n) as well. | LeetCode 1: Two Sum
two-sum
Given an array of integers `nums` and an integer `target`, return _indices of the two numbers such that they add up to `target`_.
You may assume that each input would have **_exactly_ one solution**, and you may not use the _same_ element twice.
You can return the answer in any order.
**Example 1:**
**Input:** nums = \[2,7,11,15\], target = 9
**Output:** \[0,1\]
**Explanation:** Because nums\[0\] + nums\[1\] == 9, we return \[0, 1\].
**Example 2:**
**Input:** nums = \[3,2,4\], target = 6
**Output:** \[1,2\]
**Example 3:**
**Input:** nums = \[3,3\], target = 6
**Output:** \[0,1\]
**Constraints:**
* `2 <= nums.length <= 104`
* `-109 <= nums[i] <= 109`
* `-109 <= target <= 109`
* **Only one valid answer exists.**
**Follow-up:** Can you come up with an algorithm that is less than `O(n2)` time complexity? | Solve the following problem in javascript, and explain your solution. |
```python
def twoSum(nums, target):
map = {}
for i, num in enumerate(nums):
complement = target - num
if complement in map:
return [map[complement], i]
map[num] = i
return []
```
Explanation:
The algorithm leverages a hash map (unordered_map in C++, HashMap in Java, dictionary in Python, and Map in JavaScript). It iterates through the given 'nums' array and calculates the complementary value (target - current value). If the complementary value is already in the hash map, it means that we found a solution, and we return those indices. If the complement is not in the hash map, we store the current element in the hash map with its index. If the algorithm doesn't find the solution, it returns an empty array or throws an exception (in Java).
This approach has a time complexity of O(n) and a space complexity of O(n) as well. | LeetCode 1: Two Sum
two-sum
Given an array of integers `nums` and an integer `target`, return _indices of the two numbers such that they add up to `target`_.
You may assume that each input would have **_exactly_ one solution**, and you may not use the _same_ element twice.
You can return the answer in any order.
**Example 1:**
**Input:** nums = \[2,7,11,15\], target = 9
**Output:** \[0,1\]
**Explanation:** Because nums\[0\] + nums\[1\] == 9, we return \[0, 1\].
**Example 2:**
**Input:** nums = \[3,2,4\], target = 6
**Output:** \[1,2\]
**Example 3:**
**Input:** nums = \[3,3\], target = 6
**Output:** \[0,1\]
**Constraints:**
* `2 <= nums.length <= 104`
* `-109 <= nums[i] <= 109`
* `-109 <= target <= 109`
* **Only one valid answer exists.**
**Follow-up:** Can you come up with an algorithm that is less than `O(n2)` time complexity? | Solve the following problem in python, and explain your solution. |
```cpp
struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(nullptr) {}
};
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode dummy(0);
ListNode* current = &dummy;
int carry = 0;
while (l1 || l2 || carry) {
int sum = (l1 ? l1->val : 0) + (l2 ? l2->val : 0) + carry;
carry = sum / 10;
current->next = new ListNode(sum % 10);
current = current->next;
if (l1) l1 = l1->next;
if (l2) l2 = l2->next;
}
return dummy.next;
}
```
Explanation:
1. Initialize a dummy ListNode with a value of 0.
2. Set current to that dummy ListNode, and set carry to 0.
3. Iterate over the list nodes of l1 and l2, as well as the carry, in a while loop until all are null or 0.
4. Calculate the sum of the node values and carry, store the carry for the next iteration, and store the value % 10 in a new ListNode connected to the current ListNode.
5. Shift the current ListNode, l1, and l2 to the next node if available.
6. Return the next of the dummy ListNode as a result. | LeetCode 2: Add Two Numbers
add-two-numbers
You are given two **non-empty** linked lists representing two non-negative integers. The digits are stored in **reverse order**, and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
**Example 1:**
**Input:** l1 = \[2,4,3\], l2 = \[5,6,4\]
**Output:** \[7,0,8\]
**Explanation:** 342 + 465 = 807.
**Example 2:**
**Input:** l1 = \[0\], l2 = \[0\]
**Output:** \[0\]
**Example 3:**
**Input:** l1 = \[9,9,9,9,9,9,9\], l2 = \[9,9,9,9\]
**Output:** \[8,9,9,9,0,0,0,1\]
**Constraints:**
* The number of nodes in each linked list is in the range `[1, 100]`.
* `0 <= Node.val <= 9`
* It is guaranteed that the list represents a number that does not have leading zeros. | Solve the following problem in c++, and explain your solution. |
```java
public class ListNode {
int val;
ListNode next;
ListNode(int x) { val = x; }
}
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode dummy = new ListNode(0);
ListNode current = dummy;
int carry = 0;
while (l1 != null || l2 != null || carry != 0) {
int sum = (l1 != null ? l1.val : 0) + (l2 != null ? l2.val : 0) + carry;
carry = sum / 10;
current.next = new ListNode(sum % 10);
current = current.next;
if (l1 != null) l1 = l1.next;
if (l2 != null) l2 = l2.next;
}
return dummy.next;
}
```
Explanation:
1. Initialize a dummy ListNode with a value of 0.
2. Set current to that dummy ListNode, and set carry to 0.
3. Iterate over the list nodes of l1 and l2, as well as the carry, in a while loop until all are null or 0.
4. Calculate the sum of the node values and carry, store the carry for the next iteration, and store the value % 10 in a new ListNode connected to the current ListNode.
5. Shift the current ListNode, l1, and l2 to the next node if available.
6. Return the next of the dummy ListNode as a result. | LeetCode 2: Add Two Numbers
add-two-numbers
You are given two **non-empty** linked lists representing two non-negative integers. The digits are stored in **reverse order**, and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
**Example 1:**
**Input:** l1 = \[2,4,3\], l2 = \[5,6,4\]
**Output:** \[7,0,8\]
**Explanation:** 342 + 465 = 807.
**Example 2:**
**Input:** l1 = \[0\], l2 = \[0\]
**Output:** \[0\]
**Example 3:**
**Input:** l1 = \[9,9,9,9,9,9,9\], l2 = \[9,9,9,9\]
**Output:** \[8,9,9,9,0,0,0,1\]
**Constraints:**
* The number of nodes in each linked list is in the range `[1, 100]`.
* `0 <= Node.val <= 9`
* It is guaranteed that the list represents a number that does not have leading zeros. | Solve the following problem in java, and explain your solution. |
```javascript
function ListNode(val, next) {
this.val = (val===undefined ? 0 : val);
this.next = (next===undefined ? null : next);
};
function addTwoNumbers(l1, l2) {
const dummy = new ListNode(0);
let current = dummy;
let carry = 0;
while (l1 || l2 || carry) {
let sum = (l1 ? l1.val : 0) + (l2 ? l2.val : 0) + carry;
carry = Math.floor(sum / 10);
current.next = new ListNode(sum % 10);
current = current.next;
if (l1) l1 = l1.next;
if (l2) l2 = l2.next;
}
return dummy.next;
};
```
Explanation:
1. Initialize a dummy ListNode with a value of 0.
2. Set current to that dummy ListNode, and set carry to 0.
3. Iterate over the list nodes of l1 and l2, as well as the carry, in a while loop until all are null or 0.
4. Calculate the sum of the node values and carry, store the carry for the next iteration, and store the value % 10 in a new ListNode connected to the current ListNode.
5. Shift the current ListNode, l1, and l2 to the next node if available.
6. Return the next of the dummy ListNode as a result. | LeetCode 2: Add Two Numbers
add-two-numbers
You are given two **non-empty** linked lists representing two non-negative integers. The digits are stored in **reverse order**, and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
**Example 1:**
**Input:** l1 = \[2,4,3\], l2 = \[5,6,4\]
**Output:** \[7,0,8\]
**Explanation:** 342 + 465 = 807.
**Example 2:**
**Input:** l1 = \[0\], l2 = \[0\]
**Output:** \[0\]
**Example 3:**
**Input:** l1 = \[9,9,9,9,9,9,9\], l2 = \[9,9,9,9\]
**Output:** \[8,9,9,9,0,0,0,1\]
**Constraints:**
* The number of nodes in each linked list is in the range `[1, 100]`.
* `0 <= Node.val <= 9`
* It is guaranteed that the list represents a number that does not have leading zeros. | Solve the following problem in javascript, and explain your solution. |
```python
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
def addTwoNumbers(l1, l2):
dummy = ListNode(0)
current = dummy
carry = 0
while l1 or l2 or carry:
sum_val = (l1.val if l1 else 0) + (l2.val if l2 else 0) + carry
carry = sum_val // 10
current.next = ListNode(sum_val % 10)
current = current.next
if l1: l1 = l1.next
if l2: l2 = l2.next
return dummy.next
```
Explanation:
1. Initialize a dummy ListNode with a value of 0.
2. Set current to that dummy ListNode, and set carry to 0.
3. Iterate over the list nodes of l1 and l2, as well as the carry, in a while loop until all are null or 0.
4. Calculate the sum of the node values and carry, store the carry for the next iteration, and store the value % 10 in a new ListNode connected to the current ListNode.
5. Shift the current ListNode, l1, and l2 to the next node if available.
6. Return the next of the dummy ListNode as a result. | LeetCode 2: Add Two Numbers
add-two-numbers
You are given two **non-empty** linked lists representing two non-negative integers. The digits are stored in **reverse order**, and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
**Example 1:**
**Input:** l1 = \[2,4,3\], l2 = \[5,6,4\]
**Output:** \[7,0,8\]
**Explanation:** 342 + 465 = 807.
**Example 2:**
**Input:** l1 = \[0\], l2 = \[0\]
**Output:** \[0\]
**Example 3:**
**Input:** l1 = \[9,9,9,9,9,9,9\], l2 = \[9,9,9,9\]
**Output:** \[8,9,9,9,0,0,0,1\]
**Constraints:**
* The number of nodes in each linked list is in the range `[1, 100]`.
* `0 <= Node.val <= 9`
* It is guaranteed that the list represents a number that does not have leading zeros. | Solve the following problem in python, and explain your solution. |
```cpp
#include <string>
#include <unordered_set>
int lengthOfLongestSubstring(std::string s) {
int left = 0, right = 0, maxLength = 0;
std::unordered_set<char> characters;
while (right < s.size()) {
if (characters.find(s[right]) == characters.end()) {
characters.insert(s[right]);
maxLength = std::max(maxLength, right - left + 1);
right++;
} else {
characters.erase(s[left]);
left++;
}
}
return maxLength;
}
```
Explanation:
The algorithm uses a sliding window with two pointers, left and right, to iterate through the string. It also uses a set to store the unique characters in the current window.
1. Initialize left and right pointers to the start of the string, and maxLength to 0.
2. Check if the character at the right index is in the set.
- If it's not in the set, add the character to the set, update maxLength, and move the right pointer forward.
- If it's in the set, remove the character at the left index from the set, and move the left pointer forward.
3. Repeat step 2 until the right pointer reaches the end of the string.
4. Return maxLength.
The algorithm runs in O(n) time, where n is the length of the input string. | LeetCode 3: Longest Substring Without Repeating Characters
longest-substring-without-repeating-characters
Given a string `s`, find the length of the **longest** **substring** without repeating characters.
**Example 1:**
**Input:** s = "abcabcbb "
**Output:** 3
**Explanation:** The answer is "abc ", with the length of 3.
**Example 2:**
**Input:** s = "bbbbb "
**Output:** 1
**Explanation:** The answer is "b ", with the length of 1.
**Example 3:**
**Input:** s = "pwwkew "
**Output:** 3
**Explanation:** The answer is "wke ", with the length of 3.
Notice that the answer must be a substring, "pwke " is a subsequence and not a substring.
**Constraints:**
* `0 <= s.length <= 5 * 104`
* `s` consists of English letters, digits, symbols and spaces. | Solve the following problem in c++, and explain your solution. |
```java
import java.util.HashSet;
import java.util.Set;
public int lengthOfLongestSubstring(String s) {
int left = 0, right = 0, maxLength = 0;
Set<Character> characters = new HashSet<>();
while (right < s.length()) {
if (!characters.contains(s.charAt(right))) {
characters.add(s.charAt(right));
maxLength = Math.max(maxLength, right - left + 1);
right++;
} else {
characters.remove(s.charAt(left));
left++;
}
}
return maxLength;
}
```
Explanation:
The algorithm uses a sliding window with two pointers, left and right, to iterate through the string. It also uses a set to store the unique characters in the current window.
1. Initialize left and right pointers to the start of the string, and maxLength to 0.
2. Check if the character at the right index is in the set.
- If it's not in the set, add the character to the set, update maxLength, and move the right pointer forward.
- If it's in the set, remove the character at the left index from the set, and move the left pointer forward.
3. Repeat step 2 until the right pointer reaches the end of the string.
4. Return maxLength.
The algorithm runs in O(n) time, where n is the length of the input string. | LeetCode 3: Longest Substring Without Repeating Characters
longest-substring-without-repeating-characters
Given a string `s`, find the length of the **longest** **substring** without repeating characters.
**Example 1:**
**Input:** s = "abcabcbb "
**Output:** 3
**Explanation:** The answer is "abc ", with the length of 3.
**Example 2:**
**Input:** s = "bbbbb "
**Output:** 1
**Explanation:** The answer is "b ", with the length of 1.
**Example 3:**
**Input:** s = "pwwkew "
**Output:** 3
**Explanation:** The answer is "wke ", with the length of 3.
Notice that the answer must be a substring, "pwke " is a subsequence and not a substring.
**Constraints:**
* `0 <= s.length <= 5 * 104`
* `s` consists of English letters, digits, symbols and spaces. | Solve the following problem in java, and explain your solution. |
```javascript
function lengthOfLongestSubstring(s) {
let left = 0, right = 0, maxLength = 0;
const characters = new Set();
while (right < s.length) {
if (!characters.has(s.charAt(right))) {
characters.add(s.charAt(right));
maxLength = Math.max(maxLength, right - left + 1);
right++;
} else {
characters.delete(s.charAt(left));
left++;
}
}
return maxLength;
}
```
Explanation:
The algorithm uses a sliding window with two pointers, left and right, to iterate through the string. It also uses a set to store the unique characters in the current window.
1. Initialize left and right pointers to the start of the string, and maxLength to 0.
2. Check if the character at the right index is in the set.
- If it's not in the set, add the character to the set, update maxLength, and move the right pointer forward.
- If it's in the set, remove the character at the left index from the set, and move the left pointer forward.
3. Repeat step 2 until the right pointer reaches the end of the string.
4. Return maxLength.
The algorithm runs in O(n) time, where n is the length of the input string. | LeetCode 3: Longest Substring Without Repeating Characters
longest-substring-without-repeating-characters
Given a string `s`, find the length of the **longest** **substring** without repeating characters.
**Example 1:**
**Input:** s = "abcabcbb "
**Output:** 3
**Explanation:** The answer is "abc ", with the length of 3.
**Example 2:**
**Input:** s = "bbbbb "
**Output:** 1
**Explanation:** The answer is "b ", with the length of 1.
**Example 3:**
**Input:** s = "pwwkew "
**Output:** 3
**Explanation:** The answer is "wke ", with the length of 3.
Notice that the answer must be a substring, "pwke " is a subsequence and not a substring.
**Constraints:**
* `0 <= s.length <= 5 * 104`
* `s` consists of English letters, digits, symbols and spaces. | Solve the following problem in javascript, and explain your solution. |
```python
def length_of_longest_substring(s: str) -> int:
left = 0
right = 0
max_length = 0
characters = set()
while right < len(s):
if s[right] not in characters:
characters.add(s[right])
max_length = max(max_length, right - left + 1)
right += 1
else:
characters.remove(s[left])
left += 1
return max_length
```
Explanation:
The algorithm uses a sliding window with two pointers, left and right, to iterate through the string. It also uses a set to store the unique characters in the current window.
1. Initialize left and right pointers to the start of the string, and maxLength to 0.
2. Check if the character at the right index is in the set.
- If it's not in the set, add the character to the set, update maxLength, and move the right pointer forward.
- If it's in the set, remove the character at the left index from the set, and move the left pointer forward.
3. Repeat step 2 until the right pointer reaches the end of the string.
4. Return maxLength.
The algorithm runs in O(n) time, where n is the length of the input string. | LeetCode 3: Longest Substring Without Repeating Characters
longest-substring-without-repeating-characters
Given a string `s`, find the length of the **longest** **substring** without repeating characters.
**Example 1:**
**Input:** s = "abcabcbb "
**Output:** 3
**Explanation:** The answer is "abc ", with the length of 3.
**Example 2:**
**Input:** s = "bbbbb "
**Output:** 1
**Explanation:** The answer is "b ", with the length of 1.
**Example 3:**
**Input:** s = "pwwkew "
**Output:** 3
**Explanation:** The answer is "wke ", with the length of 3.
Notice that the answer must be a substring, "pwke " is a subsequence and not a substring.
**Constraints:**
* `0 <= s.length <= 5 * 104`
* `s` consists of English letters, digits, symbols and spaces. | Solve the following problem in python, and explain your solution. |
```cpp
double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
if (nums1.size() > nums2.size()) {
return findMedianSortedArrays(nums2, nums1);
}
int x = nums1.size();
int y = nums2.size();
int low = 0;
int high = x;
while (low <= high) {
int partitionX = (low + high) / 2;
int partitionY = (x + y + 1) / 2 - partitionX;
int maxLeftX = (partitionX == 0) ? INT_MIN : nums1[partitionX - 1];
int minRightX = (partitionX == x) ? INT_MAX : nums1[partitionX];
int maxLeftY = (partitionY == 0) ? INT_MIN : nums2[partitionY - 1];
int minRightY = (partitionY == y) ? INT_MAX : nums2[partitionY];
if (maxLeftX <= minRightY && maxLeftY <= minRightX) {
if ((x + y) % 2 == 0) {
return (double(max(maxLeftX, maxLeftY) + min(minRightX, minRightY))) / 2;
} else {
return double(max(maxLeftX, maxLeftY));
}
} else if (maxLeftX > minRightY) {
high = partitionX - 1;
} else {
low = partitionX + 1;
}
}
return 0;
}
```
Explanation:
1. Choose the smaller array as nums1 so that the problem is simpler with less log(n) complexity.
2. Use Binary Search (BS) to partition the smallest array.
3. Now we calculate the position of partition in the larger array (nums2) having fetched the smaller one.
4. Find the four important numbers - maxSize - left and right of partition in the two arrays.
5. If maxSizeLeft <= minSizeRight and maxSizeLeft2 <= minSizeRight2, then the partition of both arrays is correct, if not, adjust the partition of nums1. If maxLeftX > minRightY, move the BS partition to the left; if maxLeftY > minRightX, move the BS partition to the right.
6. When the correct partition is found, calculate the median based on the length of the merged array, even or odd. | LeetCode 4: Median of Two Sorted Arrays
median-of-two-sorted-arrays
Given two sorted arrays `nums1` and `nums2` of size `m` and `n` respectively, return **the median** of the two sorted arrays.
The overall run time complexity should be `O(log (m+n))`.
**Example 1:**
**Input:** nums1 = \[1,3\], nums2 = \[2\]
**Output:** 2.00000
**Explanation:** merged array = \[1,2,3\] and median is 2.
**Example 2:**
**Input:** nums1 = \[1,2\], nums2 = \[3,4\]
**Output:** 2.50000
**Explanation:** merged array = \[1,2,3,4\] and median is (2 + 3) / 2 = 2.5.
**Constraints:**
* `nums1.length == m`
* `nums2.length == n`
* `0 <= m <= 1000`
* `0 <= n <= 1000`
* `1 <= m + n <= 2000`
* `-106 <= nums1[i], nums2[i] <= 106` | Solve the following problem in c++, and explain your solution. |
```java
public double findMedianSortedArrays(int[] nums1, int[] nums2) {
if (nums1.length > nums2.length) {
return findMedianSortedArrays(nums2, nums1);
}
int x = nums1.length;
int y = nums2.length;
int low = 0;
int high = x;
while (low <= high) {
int partitionX = (low + high) / 2;
int partitionY = (x + y + 1) / 2 - partitionX;
int maxLeftX = (partitionX == 0) ? Integer.MIN_VALUE : nums1[partitionX - 1];
int minRightX = (partitionX == x) ? Integer.MAX_VALUE : nums1[partitionX];
int maxLeftY = (partitionY == 0) ? Integer.MIN_VALUE : nums2[partitionY - 1];
int minRightY = (partitionY == y) ? Integer.MAX_VALUE : nums2[partitionY];
if (maxLeftX <= minRightY && maxLeftY <= minRightX) {
if ((x + y) % 2 == 0) {
return (double)(Math.max(maxLeftX, maxLeftY) + Math.min(minRightX, minRightY)) / 2;
} else {
return (double)Math.max(maxLeftX, maxLeftY);
}
} else if (maxLeftX > minRightY) {
high = partitionX - 1;
} else {
low = partitionX + 1;
}
}
return 0;
}
```
Explanation:
1. Choose the smaller array as nums1 so that the problem is simpler with less log(n) complexity.
2. Use Binary Search (BS) to partition the smallest array.
3. Now we calculate the position of partition in the larger array (nums2) having fetched the smaller one.
4. Find the four important numbers - maxSize - left and right of partition in the two arrays.
5. If maxSizeLeft <= minSizeRight and maxSizeLeft2 <= minSizeRight2, then the partition of both arrays is correct, if not, adjust the partition of nums1. If maxLeftX > minRightY, move the BS partition to the left; if maxLeftY > minRightX, move the BS partition to the right.
6. When the correct partition is found, calculate the median based on the length of the merged array, even or odd. | LeetCode 4: Median of Two Sorted Arrays
median-of-two-sorted-arrays
Given two sorted arrays `nums1` and `nums2` of size `m` and `n` respectively, return **the median** of the two sorted arrays.
The overall run time complexity should be `O(log (m+n))`.
**Example 1:**
**Input:** nums1 = \[1,3\], nums2 = \[2\]
**Output:** 2.00000
**Explanation:** merged array = \[1,2,3\] and median is 2.
**Example 2:**
**Input:** nums1 = \[1,2\], nums2 = \[3,4\]
**Output:** 2.50000
**Explanation:** merged array = \[1,2,3,4\] and median is (2 + 3) / 2 = 2.5.
**Constraints:**
* `nums1.length == m`
* `nums2.length == n`
* `0 <= m <= 1000`
* `0 <= n <= 1000`
* `1 <= m + n <= 2000`
* `-106 <= nums1[i], nums2[i] <= 106` | Solve the following problem in java, and explain your solution. |
```javascript
function findMedianSortedArrays(nums1, nums2) {
if (nums1.length > nums2.length) {
return findMedianSortedArrays(nums2, nums1);
}
const x = nums1.length;
const y = nums2.length;
let low = 0;
let high = x;
while (low <= high) {
const partitionX = Math.floor((low + high) / 2);
const partitionY = Math.floor((x + y + 1) / 2) - partitionX;
const maxLeftX = (partitionX === 0) ? Number.NEGATIVE_INFINITY : nums1[partitionX - 1];
const minRightX = (partitionX === x) ? Number.POSITIVE_INFINITY : nums1[partitionX];
const maxLeftY = (partitionY === 0) ? Number.NEGATIVE_INFINITY : nums2[partitionY - 1];
const minRightY = (partitionY === y) ? Number.POSITIVE_INFINITY : nums2[partitionY];
if (maxLeftX <= minRightY && maxLeftY <= minRightX) {
if ((x + y) % 2 === 0) {
return (Math.max(maxLeftX, maxLeftY) + Math.min(minRightX, minRightY)) / 2;
} else {
return Math.max(maxLeftX, maxLeftY);
}
} else if (maxLeftX > minRightY) {
high = partitionX - 1;
} else {
low = partitionX + 1;
}
}
return 0;
}
```
Explanation:
1. Choose the smaller array as nums1 so that the problem is simpler with less log(n) complexity.
2. Use Binary Search (BS) to partition the smallest array.
3. Now we calculate the position of partition in the larger array (nums2) having fetched the smaller one.
4. Find the four important numbers - maxSize - left and right of partition in the two arrays.
5. If maxSizeLeft <= minSizeRight and maxSizeLeft2 <= minSizeRight2, then the partition of both arrays is correct, if not, adjust the partition of nums1. If maxLeftX > minRightY, move the BS partition to the left; if maxLeftY > minRightX, move the BS partition to the right.
6. When the correct partition is found, calculate the median based on the length of the merged array, even or odd. | LeetCode 4: Median of Two Sorted Arrays
median-of-two-sorted-arrays
Given two sorted arrays `nums1` and `nums2` of size `m` and `n` respectively, return **the median** of the two sorted arrays.
The overall run time complexity should be `O(log (m+n))`.
**Example 1:**
**Input:** nums1 = \[1,3\], nums2 = \[2\]
**Output:** 2.00000
**Explanation:** merged array = \[1,2,3\] and median is 2.
**Example 2:**
**Input:** nums1 = \[1,2\], nums2 = \[3,4\]
**Output:** 2.50000
**Explanation:** merged array = \[1,2,3,4\] and median is (2 + 3) / 2 = 2.5.
**Constraints:**
* `nums1.length == m`
* `nums2.length == n`
* `0 <= m <= 1000`
* `0 <= n <= 1000`
* `1 <= m + n <= 2000`
* `-106 <= nums1[i], nums2[i] <= 106` | Solve the following problem in javascript, and explain your solution. |
```python
def findMedianSortedArrays(nums1, nums2):
if len(nums1) > len(nums2):
return findMedianSortedArrays(nums2, nums1)
x, y = len(nums1), len(nums2)
low, high = 0, x
while low <= high:
partition_x = (low + high) // 2
partition_y = (x + y + 1) // 2 - partition_x
max_left_x = float('-inf') if partition_x == 0 else nums1[partition_x - 1]
min_right_x = float('inf') if partition_x == x else nums1[partition_x]
max_left_y = float('-inf') if partition_y == 0 else nums2[partition_y - 1]
min_right_y = float('inf') if partition_y == y else nums2[partition_y]
if max_left_x <= min_right_y and max_left_y <= min_right_x:
if (x + y) % 2 == 0:
return (max(max_left_x, max_left_y) + min(min_right_x, min_right_y)) / 2
else:
return max(max_left_x, max_left_y)
elif max_left_x > min_right_y:
high = partition_x - 1
else:
low = partition_x + 1
return 0
```
Explanation:
1. Choose the smaller array as nums1 so that the problem is simpler with less log(n) complexity.
2. Use Binary Search (BS) to partition the smallest array.
3. Now we calculate the position of partition in the larger array (nums2) having fetched the smaller one.
4. Find the four important numbers - maxSize - left and right of partition in the two arrays.
5. If maxSizeLeft <= minSizeRight and maxSizeLeft2 <= minSizeRight2, then the partition of both arrays is correct, if not, adjust the partition of nums1. If maxLeftX > minRightY, move the BS partition to the left; if maxLeftY > minRightX, move the BS partition to the right.
6. When the correct partition is found, calculate the median based on the length of the merged array, even or odd. | LeetCode 4: Median of Two Sorted Arrays
median-of-two-sorted-arrays
Given two sorted arrays `nums1` and `nums2` of size `m` and `n` respectively, return **the median** of the two sorted arrays.
The overall run time complexity should be `O(log (m+n))`.
**Example 1:**
**Input:** nums1 = \[1,3\], nums2 = \[2\]
**Output:** 2.00000
**Explanation:** merged array = \[1,2,3\] and median is 2.
**Example 2:**
**Input:** nums1 = \[1,2\], nums2 = \[3,4\]
**Output:** 2.50000
**Explanation:** merged array = \[1,2,3,4\] and median is (2 + 3) / 2 = 2.5.
**Constraints:**
* `nums1.length == m`
* `nums2.length == n`
* `0 <= m <= 1000`
* `0 <= n <= 1000`
* `1 <= m + n <= 2000`
* `-106 <= nums1[i], nums2[i] <= 106` | Solve the following problem in python, and explain your solution. |
```cpp
#include <string>
std::string longestPalindromicSubstring(std::string s) {
int n = s.size();
if (n == 0) return "";
int start = 0, maxLength = 1;
for (int i = 0; i < n; ++i) {
int l = i, r = i;
while (r < n - 1 && s[r] == s[r + 1])
r++;
i = r;
while (l > 0 && r < n - 1 && s[l - 1] == s[r + 1]) {
l--;
r++;
}
int length = r - l + 1;
if (length > maxLength) {
start = l;
maxLength = length;
}
}
return s.substr(start, maxLength);
}
```
Explanation:
1. Initialize `start` and `maxLength` for result substring.
2. Iterate through the given string `s` using the index `i`.
3. For each index `i`, create two pointers `l` and `r` starting at `i`.
4. Check if there's a consecutive sequence of identical characters, increment the right pointer `r` until the end of the sequence is reached.
5. Update the index `i` to the current value of `r`.
6. Expand the pointers `l` and `r` outwards to find the longest palindromic substring, checking that characters on both sides are equal.
7. If the current length of the substring is greater than `maxLength`, update `start` and `maxLength`.
8. Return the longest palindromic substring using the `start` and `maxLength`. | LeetCode 5: Longest Palindromic Substring
longest-palindromic-substring
Given a string `s`, return _the longest_ _palindromic_ _substring_ in `s`.
**Example 1:**
**Input:** s = "babad "
**Output:** "bab "
**Explanation:** "aba " is also a valid answer.
**Example 2:**
**Input:** s = "cbbd "
**Output:** "bb "
**Constraints:**
* `1 <= s.length <= 1000`
* `s` consist of only digits and English letters. | Solve the following problem in c++, and explain your solution. |
```java
public String longestPalindromicSubstring(String s) {
int n = s.length();
if (n == 0) return "";
int start = 0, maxLength = 1;
for (int i = 0; i < n; ++i) {
int l = i, r = i;
while (r < n - 1 && s.charAt(r) == s.charAt(r + 1))
r++;
i = r;
while (l > 0 && r < n - 1 && s.charAt(l - 1) == s.charAt(r + 1)) {
l--;
r++;
}
int length = r - l + 1;
if (length > maxLength) {
start = l;
maxLength = length;
}
}
return s.substring(start, start + maxLength);
}
```
Explanation:
1. Initialize `start` and `maxLength` for result substring.
2. Iterate through the given string `s` using the index `i`.
3. For each index `i`, create two pointers `l` and `r` starting at `i`.
4. Check if there's a consecutive sequence of identical characters, increment the right pointer `r` until the end of the sequence is reached.
5. Update the index `i` to the current value of `r`.
6. Expand the pointers `l` and `r` outwards to find the longest palindromic substring, checking that characters on both sides are equal.
7. If the current length of the substring is greater than `maxLength`, update `start` and `maxLength`.
8. Return the longest palindromic substring using the `start` and `maxLength`. | LeetCode 5: Longest Palindromic Substring
longest-palindromic-substring
Given a string `s`, return _the longest_ _palindromic_ _substring_ in `s`.
**Example 1:**
**Input:** s = "babad "
**Output:** "bab "
**Explanation:** "aba " is also a valid answer.
**Example 2:**
**Input:** s = "cbbd "
**Output:** "bb "
**Constraints:**
* `1 <= s.length <= 1000`
* `s` consist of only digits and English letters. | Solve the following problem in java, and explain your solution. |
```javascript
function longestPalindromicSubstring(s) {
let n = s.length;
if (n === 0) return "";
let start = 0, maxLength = 1;
for (let i = 0; i < n; ++i) {
let l = i, r = i;
while (r < n - 1 && s[r] === s[r + 1])
r++;
i = r;
while (l > 0 && r < n - 1 && s[l - 1] === s[r + 1]) {
l--;
r++;
}
let length = r - l + 1;
if (length > maxLength) {
start = l;
maxLength = length;
}
}
return s.substring(start, start + maxLength);
}
```
Explanation:
1. Initialize `start` and `maxLength` for result substring.
2. Iterate through the given string `s` using the index `i`.
3. For each index `i`, create two pointers `l` and `r` starting at `i`.
4. Check if there's a consecutive sequence of identical characters, increment the right pointer `r` until the end of the sequence is reached.
5. Update the index `i` to the current value of `r`.
6. Expand the pointers `l` and `r` outwards to find the longest palindromic substring, checking that characters on both sides are equal.
7. If the current length of the substring is greater than `maxLength`, update `start` and `maxLength`.
8. Return the longest palindromic substring using the `start` and `maxLength`. | LeetCode 5: Longest Palindromic Substring
longest-palindromic-substring
Given a string `s`, return _the longest_ _palindromic_ _substring_ in `s`.
**Example 1:**
**Input:** s = "babad "
**Output:** "bab "
**Explanation:** "aba " is also a valid answer.
**Example 2:**
**Input:** s = "cbbd "
**Output:** "bb "
**Constraints:**
* `1 <= s.length <= 1000`
* `s` consist of only digits and English letters. | Solve the following problem in javascript, and explain your solution. |
```python
def longest_palindromic_substring(s: str) -> str:
n = len(s)
if n == 0: return ""
start, max_length = 0, 1
for i in range(n):
l, r = i, i
while r < n - 1 and s[r] == s[r + 1]:
r += 1
i = r
while l > 0 and r < n - 1 and s[l - 1] == s[r + 1]:
l -= 1
r += 1
length = r - l + 1
if length > max_length:
start, max_length = l, length
return s[start:start + max_length]
```
Explanation:
1. Initialize `start` and `maxLength` for result substring.
2. Iterate through the given string `s` using the index `i`.
3. For each index `i`, create two pointers `l` and `r` starting at `i`.
4. Check if there's a consecutive sequence of identical characters, increment the right pointer `r` until the end of the sequence is reached.
5. Update the index `i` to the current value of `r`.
6. Expand the pointers `l` and `r` outwards to find the longest palindromic substring, checking that characters on both sides are equal.
7. If the current length of the substring is greater than `maxLength`, update `start` and `maxLength`.
8. Return the longest palindromic substring using the `start` and `maxLength`. | LeetCode 5: Longest Palindromic Substring
longest-palindromic-substring
Given a string `s`, return _the longest_ _palindromic_ _substring_ in `s`.
**Example 1:**
**Input:** s = "babad "
**Output:** "bab "
**Explanation:** "aba " is also a valid answer.
**Example 2:**
**Input:** s = "cbbd "
**Output:** "bb "
**Constraints:**
* `1 <= s.length <= 1000`
* `s` consist of only digits and English letters. | Solve the following problem in python, and explain your solution. |
```cpp
#include <string>
#include <vector>
std::string convert(std::string s, int numRows) {
if (numRows == 1 || numRows >= s.size()) return s;
std::vector<std::string> rows(std::min(numRows, int(s.size())));
int curRow = 0;
bool goingDown = false;
for (char c : s) {
rows[curRow] += c;
if (curRow == 0 || curRow == numRows - 1) goingDown = !goingDown;
curRow += goingDown ? 1 : -1;
}
std::string result;
for (std::string row : rows) result += row;
return result;
}
```
Explanation:
1. If the number of rows is 1 or is greater than or equal to the length of the string, the string is just returned as is.
2. Create an array called "rows" to contain the characters that would reside in each row of the zigzag pattern.
3. Iterate through the string one character at a time.
4. For each character in the string, place it into the appropriate row of the "rows" array.
5. Determine whether the next character should be placed in the row above or below the current row by checking if the current row is 0 (at the top) or equal to the number of rows minus 1 (at the bottom). If at the top or the bottom, the direction will change.
6. Update the current row index according to the direction (going up or down).
7. After iterating through the entire string, join the rows array into a single string and return it. | LeetCode 6: Zigzag Conversion
zigzag-conversion
The string `"PAYPALISHIRING "` is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)
P A H N
A P L S I I G
Y I R
And then read line by line: `"PAHNAPLSIIGYIR "`
Write the code that will take a string and make this conversion given a number of rows:
string convert(string s, int numRows);
**Example 1:**
**Input:** s = "PAYPALISHIRING ", numRows = 3
**Output:** "PAHNAPLSIIGYIR "
**Example 2:**
**Input:** s = "PAYPALISHIRING ", numRows = 4
**Output:** "PINALSIGYAHRPI "
**Explanation:**
P I N
A L S I G
Y A H R
P I
**Example 3:**
**Input:** s = "A ", numRows = 1
**Output:** "A "
**Constraints:**
* `1 <= s.length <= 1000`
* `s` consists of English letters (lower-case and upper-case), `','` and `'.'`.
* `1 <= numRows <= 1000` | Solve the following problem in c++, and explain your solution. |
```java
public class ZigzagConversion {
public String convert(String s, int numRows) {
if (numRows == 1 || numRows >= s.length()) return s;
StringBuilder[] rows = new StringBuilder[numRows];
for (int i = 0; i < numRows; i++) rows[i] = new StringBuilder();
int curRow = 0;
boolean goingDown = false;
for (char c : s.toCharArray()) {
rows[curRow].append(c);
if (curRow == 0 || curRow == numRows - 1) goingDown = !goingDown;
curRow += goingDown ? 1 : -1;
}
StringBuilder result = new StringBuilder();
for (StringBuilder row : rows) result.append(row);
return result.toString();
}
}
```
Explanation:
1. If the number of rows is 1 or is greater than or equal to the length of the string, the string is just returned as is.
2. Create an array called "rows" to contain the characters that would reside in each row of the zigzag pattern.
3. Iterate through the string one character at a time.
4. For each character in the string, place it into the appropriate row of the "rows" array.
5. Determine whether the next character should be placed in the row above or below the current row by checking if the current row is 0 (at the top) or equal to the number of rows minus 1 (at the bottom). If at the top or the bottom, the direction will change.
6. Update the current row index according to the direction (going up or down).
7. After iterating through the entire string, join the rows array into a single string and return it. | LeetCode 6: Zigzag Conversion
zigzag-conversion
The string `"PAYPALISHIRING "` is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)
P A H N
A P L S I I G
Y I R
And then read line by line: `"PAHNAPLSIIGYIR "`
Write the code that will take a string and make this conversion given a number of rows:
string convert(string s, int numRows);
**Example 1:**
**Input:** s = "PAYPALISHIRING ", numRows = 3
**Output:** "PAHNAPLSIIGYIR "
**Example 2:**
**Input:** s = "PAYPALISHIRING ", numRows = 4
**Output:** "PINALSIGYAHRPI "
**Explanation:**
P I N
A L S I G
Y A H R
P I
**Example 3:**
**Input:** s = "A ", numRows = 1
**Output:** "A "
**Constraints:**
* `1 <= s.length <= 1000`
* `s` consists of English letters (lower-case and upper-case), `','` and `'.'`.
* `1 <= numRows <= 1000` | Solve the following problem in java, and explain your solution. |
```javascript
function convert(s, numRows) {
if (numRows === 1 || numRows >= s.length) return s;
let rows = new Array(numRows).fill("");
let curRow = 0;
let goingDown = false;
for (let c of s) {
rows[curRow] += c;
if (curRow === 0 || curRow === numRows - 1) goingDown = !goingDown;
curRow += goingDown ? 1 : -1;
}
return rows.join("");
}
```
Explanation:
1. If the number of rows is 1 or is greater than or equal to the length of the string, the string is just returned as is.
2. Create an array called "rows" to contain the characters that would reside in each row of the zigzag pattern.
3. Iterate through the string one character at a time.
4. For each character in the string, place it into the appropriate row of the "rows" array.
5. Determine whether the next character should be placed in the row above or below the current row by checking if the current row is 0 (at the top) or equal to the number of rows minus 1 (at the bottom). If at the top or the bottom, the direction will change.
6. Update the current row index according to the direction (going up or down).
7. After iterating through the entire string, join the rows array into a single string and return it. | LeetCode 6: Zigzag Conversion
zigzag-conversion
The string `"PAYPALISHIRING "` is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)
P A H N
A P L S I I G
Y I R
And then read line by line: `"PAHNAPLSIIGYIR "`
Write the code that will take a string and make this conversion given a number of rows:
string convert(string s, int numRows);
**Example 1:**
**Input:** s = "PAYPALISHIRING ", numRows = 3
**Output:** "PAHNAPLSIIGYIR "
**Example 2:**
**Input:** s = "PAYPALISHIRING ", numRows = 4
**Output:** "PINALSIGYAHRPI "
**Explanation:**
P I N
A L S I G
Y A H R
P I
**Example 3:**
**Input:** s = "A ", numRows = 1
**Output:** "A "
**Constraints:**
* `1 <= s.length <= 1000`
* `s` consists of English letters (lower-case and upper-case), `','` and `'.'`.
* `1 <= numRows <= 1000` | Solve the following problem in javascript, and explain your solution. |
```python
def convert(s: str, numRows: int) -> str:
if numRows == 1 or numRows >= len(s):
return s
rows = [""] * numRows
curRow = 0
goingDown = False
for c in s:
rows[curRow] += c
if curRow == 0 or curRow == numRows - 1:
goingDown = not goingDown
curRow += 1 if goingDown else -1
return "".join(rows)
```
Explanation:
1. If the number of rows is 1 or is greater than or equal to the length of the string, the string is just returned as is.
2. Create an array called "rows" to contain the characters that would reside in each row of the zigzag pattern.
3. Iterate through the string one character at a time.
4. For each character in the string, place it into the appropriate row of the "rows" array.
5. Determine whether the next character should be placed in the row above or below the current row by checking if the current row is 0 (at the top) or equal to the number of rows minus 1 (at the bottom). If at the top or the bottom, the direction will change.
6. Update the current row index according to the direction (going up or down).
7. After iterating through the entire string, join the rows array into a single string and return it. | LeetCode 6: Zigzag Conversion
zigzag-conversion
The string `"PAYPALISHIRING "` is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)
P A H N
A P L S I I G
Y I R
And then read line by line: `"PAHNAPLSIIGYIR "`
Write the code that will take a string and make this conversion given a number of rows:
string convert(string s, int numRows);
**Example 1:**
**Input:** s = "PAYPALISHIRING ", numRows = 3
**Output:** "PAHNAPLSIIGYIR "
**Example 2:**
**Input:** s = "PAYPALISHIRING ", numRows = 4
**Output:** "PINALSIGYAHRPI "
**Explanation:**
P I N
A L S I G
Y A H R
P I
**Example 3:**
**Input:** s = "A ", numRows = 1
**Output:** "A "
**Constraints:**
* `1 <= s.length <= 1000`
* `s` consists of English letters (lower-case and upper-case), `','` and `'.'`.
* `1 <= numRows <= 1000` | Solve the following problem in python, and explain your solution. |
```cpp
int reverse(int x) {
long long res = 0;
while (x != 0) {
res = res * 10 + x % 10;
x /= 10;
}
return (res < INT_MIN || res > INT_MAX) ? 0 : res;
}
```
Explanation:
The algorithm initializes `res` as 0. Then, we determine the sign of the input integer and take its absolute value. We use a while loop to iterate through each digit of `x` from right to left. In each iteration, we multiply the current `res` by 10 and add the last digit of `x` to it. The last digit of `x` is obtained by `x % 10`. After adding the last digit, we remove it from `x` by doing either `x /= 10` or `x //= 10`.
After processing all the digits, we adjust `res` with the sign we computed earlier. Finally, we check if the reversed integer is within the 32-bit integer range. If it is, we return the result; otherwise, we return 0. | LeetCode 7: Reverse Integer
reverse-integer
Given a signed 32-bit integer `x`, return `x` _with its digits reversed_. If reversing `x` causes the value to go outside the signed 32-bit integer range `[-231, 231 - 1]`, then return `0`.
**Assume the environment does not allow you to store 64-bit integers (signed or unsigned).**
**Example 1:**
**Input:** x = 123
**Output:** 321
**Example 2:**
**Input:** x = -123
**Output:** -321
**Example 3:**
**Input:** x = 120
**Output:** 21
**Constraints:**
* `-231 <= x <= 231 - 1` | Solve the following problem in c++, and explain your solution. |
```java
public int reverse(int x) {
long res = 0;
while (x != 0) {
res = res * 10 + x % 10;
x /= 10;
}
return (res < Integer.MIN_VALUE || res > Integer.MAX_VALUE) ? 0 : (int)res;
}
```
Explanation:
The algorithm initializes `res` as 0. Then, we determine the sign of the input integer and take its absolute value. We use a while loop to iterate through each digit of `x` from right to left. In each iteration, we multiply the current `res` by 10 and add the last digit of `x` to it. The last digit of `x` is obtained by `x % 10`. After adding the last digit, we remove it from `x` by doing either `x /= 10` or `x //= 10`.
After processing all the digits, we adjust `res` with the sign we computed earlier. Finally, we check if the reversed integer is within the 32-bit integer range. If it is, we return the result; otherwise, we return 0. | LeetCode 7: Reverse Integer
reverse-integer
Given a signed 32-bit integer `x`, return `x` _with its digits reversed_. If reversing `x` causes the value to go outside the signed 32-bit integer range `[-231, 231 - 1]`, then return `0`.
**Assume the environment does not allow you to store 64-bit integers (signed or unsigned).**
**Example 1:**
**Input:** x = 123
**Output:** 321
**Example 2:**
**Input:** x = -123
**Output:** -321
**Example 3:**
**Input:** x = 120
**Output:** 21
**Constraints:**
* `-231 <= x <= 231 - 1` | Solve the following problem in java, and explain your solution. |
```javascript
function reverse(x) {
let sign = x < 0 ? -1 : 1;
x = Math.abs(x);
let res = 0;
while (x !== 0) {
res = res * 10 + x % 10;
x = Math.floor(x / 10);
}
res *= sign;
return (res < -(2 ** 31) || res > 2 ** 31 - 1) ? 0 : res;
}
```
Explanation:
The algorithm initializes `res` as 0. Then, we determine the sign of the input integer and take its absolute value. We use a while loop to iterate through each digit of `x` from right to left. In each iteration, we multiply the current `res` by 10 and add the last digit of `x` to it. The last digit of `x` is obtained by `x % 10`. After adding the last digit, we remove it from `x` by doing either `x /= 10` or `x //= 10`.
After processing all the digits, we adjust `res` with the sign we computed earlier. Finally, we check if the reversed integer is within the 32-bit integer range. If it is, we return the result; otherwise, we return 0. | LeetCode 7: Reverse Integer
reverse-integer
Given a signed 32-bit integer `x`, return `x` _with its digits reversed_. If reversing `x` causes the value to go outside the signed 32-bit integer range `[-231, 231 - 1]`, then return `0`.
**Assume the environment does not allow you to store 64-bit integers (signed or unsigned).**
**Example 1:**
**Input:** x = 123
**Output:** 321
**Example 2:**
**Input:** x = -123
**Output:** -321
**Example 3:**
**Input:** x = 120
**Output:** 21
**Constraints:**
* `-231 <= x <= 231 - 1` | Solve the following problem in javascript, and explain your solution. |
```python
def reverse(x: int) -> int:
sign = -1 if x < 0 else 1
x = abs(x)
res = 0
while x:
res = res * 10 + x % 10
x //= 10
res *= sign
return res if -2**31 <= res <= 2**31 - 1 else 0
```
Explanation:
The algorithm initializes `res` as 0. Then, we determine the sign of the input integer and take its absolute value. We use a while loop to iterate through each digit of `x` from right to left. In each iteration, we multiply the current `res` by 10 and add the last digit of `x` to it. The last digit of `x` is obtained by `x % 10`. After adding the last digit, we remove it from `x` by doing either `x /= 10` or `x //= 10`.
After processing all the digits, we adjust `res` with the sign we computed earlier. Finally, we check if the reversed integer is within the 32-bit integer range. If it is, we return the result; otherwise, we return 0. | LeetCode 7: Reverse Integer
reverse-integer
Given a signed 32-bit integer `x`, return `x` _with its digits reversed_. If reversing `x` causes the value to go outside the signed 32-bit integer range `[-231, 231 - 1]`, then return `0`.
**Assume the environment does not allow you to store 64-bit integers (signed or unsigned).**
**Example 1:**
**Input:** x = 123
**Output:** 321
**Example 2:**
**Input:** x = -123
**Output:** -321
**Example 3:**
**Input:** x = 120
**Output:** 21
**Constraints:**
* `-231 <= x <= 231 - 1` | Solve the following problem in python, and explain your solution. |
```cpp
int myAtoi(string s) {
long result = 0;
int i = 0, sign = 1;
while (s[i] == ' ') i++;
if (s[i] == '-' || s[i] == '+') sign = (s[i++] == '+') ? 1 : -1;
while (isdigit(s[i])) {
result = result * 10 + (s[i++] - '0');
if (result * sign > INT_MAX) return INT_MAX;
if (result * sign < INT_MIN) return INT_MIN;
}
return result * sign;
}
```
Explanation:
1. Initialize `result` as 0, index `i` as 0, and `sign` as 1 (positive).
2. Loop through the string until a non-space character is found or the end of the string is reached.
3. If the current character is '+' or '-', update the `sign` according to the character and increment the index `i`.
4. If a digit is found, multiply the previous `result` by 10 and add the digit to it. Increment the index `i`.
5. Check if the result multiplied by the sign is out of the 32-bit signed integer range. If so, return the appropriate boundary value.
6. When no more digits are found, return the final result multiplied by the sign. | LeetCode 8: String to Integer (atoi)
string-to-integer-atoi
Implement the `myAtoi(string s)` function, which converts a string to a 32-bit signed integer (similar to C/C++'s `atoi` function).
The algorithm for `myAtoi(string s)` is as follows:
1. Read in and ignore any leading whitespace.
2. Check if the next character (if not already at the end of the string) is `'-'` or `'+'`. Read this character in if it is either. This determines if the final result is negative or positive respectively. Assume the result is positive if neither is present.
3. Read in next the characters until the next non-digit character or the end of the input is reached. The rest of the string is ignored.
4. Convert these digits into an integer (i.e. `"123 " -> 123`, `"0032 " -> 32`). If no digits were read, then the integer is `0`. Change the sign as necessary (from step 2).
5. If the integer is out of the 32-bit signed integer range `[-231, 231 - 1]`, then clamp the integer so that it remains in the range. Specifically, integers less than `-231` should be clamped to `-231`, and integers greater than `231 - 1` should be clamped to `231 - 1`.
6. Return the integer as the final result.
**Note:**
* Only the space character `' '` is considered a whitespace character.
* **Do not ignore** any characters other than the leading whitespace or the rest of the string after the digits.
**Example 1:**
**Input:** s = "42 "
**Output:** 42
**Explanation:** The underlined characters are what is read in, the caret is the current reader position.
Step 1: "42 " (no characters read because there is no leading whitespace)
^
Step 2: "42 " (no characters read because there is neither a '-' nor '+')
^
Step 3: "42 " ( "42 " is read in)
^
The parsed integer is 42.
Since 42 is in the range \[-231, 231 - 1\], the final result is 42.
**Example 2:**
**Input:** s = " -42 "
**Output:** -42
**Explanation:**
Step 1: " \-42 " (leading whitespace is read and ignored)
^
Step 2: " \-42 " ('-' is read, so the result should be negative)
^
Step 3: " -42 " ( "42 " is read in)
^
The parsed integer is -42.
Since -42 is in the range \[-231, 231 - 1\], the final result is -42.
**Example 3:**
**Input:** s = "4193 with words "
**Output:** 4193
**Explanation:**
Step 1: "4193 with words " (no characters read because there is no leading whitespace)
^
Step 2: "4193 with words " (no characters read because there is neither a '-' nor '+')
^
Step 3: "4193 with words " ( "4193 " is read in; reading stops because the next character is a non-digit)
^
The parsed integer is 4193.
Since 4193 is in the range \[-231, 231 - 1\], the final result is 4193.
**Constraints:**
* `0 <= s.length <= 200`
* `s` consists of English letters (lower-case and upper-case), digits (`0-9`), `' '`, `'+'`, `'-'`, and `'.'`. | Solve the following problem in c++, and explain your solution. |
```java
public int myAtoi(String s) {
long result = 0;
int i = 0, sign = 1;
while (i < s.length() && s.charAt(i) == ' ') i++;
if (i < s.length() && (s.charAt(i) == '-' || s.charAt(i) == '+')) {
sign = (s.charAt(i++) == '+') ? 1 : -1;
}
while (i < s.length() && Character.isDigit(s.charAt(i))) {
result = result * 10 + (s.charAt(i++) - '0');
if (result * sign > Integer.MAX_VALUE) return Integer.MAX_VALUE;
if (result * sign < Integer.MIN_VALUE) return Integer.MIN_VALUE;
}
return (int) (result * sign);
}
```
Explanation:
1. Initialize `result` as 0, index `i` as 0, and `sign` as 1 (positive).
2. Loop through the string until a non-space character is found or the end of the string is reached.
3. If the current character is '+' or '-', update the `sign` according to the character and increment the index `i`.
4. If a digit is found, multiply the previous `result` by 10 and add the digit to it. Increment the index `i`.
5. Check if the result multiplied by the sign is out of the 32-bit signed integer range. If so, return the appropriate boundary value.
6. When no more digits are found, return the final result multiplied by the sign. | LeetCode 8: String to Integer (atoi)
string-to-integer-atoi
Implement the `myAtoi(string s)` function, which converts a string to a 32-bit signed integer (similar to C/C++'s `atoi` function).
The algorithm for `myAtoi(string s)` is as follows:
1. Read in and ignore any leading whitespace.
2. Check if the next character (if not already at the end of the string) is `'-'` or `'+'`. Read this character in if it is either. This determines if the final result is negative or positive respectively. Assume the result is positive if neither is present.
3. Read in next the characters until the next non-digit character or the end of the input is reached. The rest of the string is ignored.
4. Convert these digits into an integer (i.e. `"123 " -> 123`, `"0032 " -> 32`). If no digits were read, then the integer is `0`. Change the sign as necessary (from step 2).
5. If the integer is out of the 32-bit signed integer range `[-231, 231 - 1]`, then clamp the integer so that it remains in the range. Specifically, integers less than `-231` should be clamped to `-231`, and integers greater than `231 - 1` should be clamped to `231 - 1`.
6. Return the integer as the final result.
**Note:**
* Only the space character `' '` is considered a whitespace character.
* **Do not ignore** any characters other than the leading whitespace or the rest of the string after the digits.
**Example 1:**
**Input:** s = "42 "
**Output:** 42
**Explanation:** The underlined characters are what is read in, the caret is the current reader position.
Step 1: "42 " (no characters read because there is no leading whitespace)
^
Step 2: "42 " (no characters read because there is neither a '-' nor '+')
^
Step 3: "42 " ( "42 " is read in)
^
The parsed integer is 42.
Since 42 is in the range \[-231, 231 - 1\], the final result is 42.
**Example 2:**
**Input:** s = " -42 "
**Output:** -42
**Explanation:**
Step 1: " \-42 " (leading whitespace is read and ignored)
^
Step 2: " \-42 " ('-' is read, so the result should be negative)
^
Step 3: " -42 " ( "42 " is read in)
^
The parsed integer is -42.
Since -42 is in the range \[-231, 231 - 1\], the final result is -42.
**Example 3:**
**Input:** s = "4193 with words "
**Output:** 4193
**Explanation:**
Step 1: "4193 with words " (no characters read because there is no leading whitespace)
^
Step 2: "4193 with words " (no characters read because there is neither a '-' nor '+')
^
Step 3: "4193 with words " ( "4193 " is read in; reading stops because the next character is a non-digit)
^
The parsed integer is 4193.
Since 4193 is in the range \[-231, 231 - 1\], the final result is 4193.
**Constraints:**
* `0 <= s.length <= 200`
* `s` consists of English letters (lower-case and upper-case), digits (`0-9`), `' '`, `'+'`, `'-'`, and `'.'`. | Solve the following problem in java, and explain your solution. |
```javascript
function myAtoi(s) {
let result = 0, i = 0, sign = 1;
while (s[i] === ' ') i++;
if (s[i] === '-' || s[i] === '+') sign = (s[i++] === '+') ? 1 : -1;
while (!isNaN(s[i]) && s[i] !== ' ') {
result = result * 10 + parseInt(s[i++], 10);
if (result * sign > 2147483647) return 2147483647;
if (result * sign < -2147483648) return -2147483648;
}
return result * sign;
}
```
Explanation:
1. Initialize `result` as 0, index `i` as 0, and `sign` as 1 (positive).
2. Loop through the string until a non-space character is found or the end of the string is reached.
3. If the current character is '+' or '-', update the `sign` according to the character and increment the index `i`.
4. If a digit is found, multiply the previous `result` by 10 and add the digit to it. Increment the index `i`.
5. Check if the result multiplied by the sign is out of the 32-bit signed integer range. If so, return the appropriate boundary value.
6. When no more digits are found, return the final result multiplied by the sign. | LeetCode 8: String to Integer (atoi)
string-to-integer-atoi
Implement the `myAtoi(string s)` function, which converts a string to a 32-bit signed integer (similar to C/C++'s `atoi` function).
The algorithm for `myAtoi(string s)` is as follows:
1. Read in and ignore any leading whitespace.
2. Check if the next character (if not already at the end of the string) is `'-'` or `'+'`. Read this character in if it is either. This determines if the final result is negative or positive respectively. Assume the result is positive if neither is present.
3. Read in next the characters until the next non-digit character or the end of the input is reached. The rest of the string is ignored.
4. Convert these digits into an integer (i.e. `"123 " -> 123`, `"0032 " -> 32`). If no digits were read, then the integer is `0`. Change the sign as necessary (from step 2).
5. If the integer is out of the 32-bit signed integer range `[-231, 231 - 1]`, then clamp the integer so that it remains in the range. Specifically, integers less than `-231` should be clamped to `-231`, and integers greater than `231 - 1` should be clamped to `231 - 1`.
6. Return the integer as the final result.
**Note:**
* Only the space character `' '` is considered a whitespace character.
* **Do not ignore** any characters other than the leading whitespace or the rest of the string after the digits.
**Example 1:**
**Input:** s = "42 "
**Output:** 42
**Explanation:** The underlined characters are what is read in, the caret is the current reader position.
Step 1: "42 " (no characters read because there is no leading whitespace)
^
Step 2: "42 " (no characters read because there is neither a '-' nor '+')
^
Step 3: "42 " ( "42 " is read in)
^
The parsed integer is 42.
Since 42 is in the range \[-231, 231 - 1\], the final result is 42.
**Example 2:**
**Input:** s = " -42 "
**Output:** -42
**Explanation:**
Step 1: " \-42 " (leading whitespace is read and ignored)
^
Step 2: " \-42 " ('-' is read, so the result should be negative)
^
Step 3: " -42 " ( "42 " is read in)
^
The parsed integer is -42.
Since -42 is in the range \[-231, 231 - 1\], the final result is -42.
**Example 3:**
**Input:** s = "4193 with words "
**Output:** 4193
**Explanation:**
Step 1: "4193 with words " (no characters read because there is no leading whitespace)
^
Step 2: "4193 with words " (no characters read because there is neither a '-' nor '+')
^
Step 3: "4193 with words " ( "4193 " is read in; reading stops because the next character is a non-digit)
^
The parsed integer is 4193.
Since 4193 is in the range \[-231, 231 - 1\], the final result is 4193.
**Constraints:**
* `0 <= s.length <= 200`
* `s` consists of English letters (lower-case and upper-case), digits (`0-9`), `' '`, `'+'`, `'-'`, and `'.'`. | Solve the following problem in javascript, and explain your solution. |
```python
def myAtoi(s):
result, i, sign = 0, 0, 1
while i < len(s) and s[i] == ' ': i += 1
if i < len(s) and (s[i] == '-' or s[i] == '+'):
sign = -1 if s[i] == '-' else 1
i += 1
while i < len(s) and s[i].isdigit():
result = result * 10 + int(s[i])
i += 1
if result * sign > 2**31 - 1: return 2**31 - 1
if result * sign < -2**31: return -2**31
return result * sign
```
Explanation:
1. Initialize `result` as 0, index `i` as 0, and `sign` as 1 (positive).
2. Loop through the string until a non-space character is found or the end of the string is reached.
3. If the current character is '+' or '-', update the `sign` according to the character and increment the index `i`.
4. If a digit is found, multiply the previous `result` by 10 and add the digit to it. Increment the index `i`.
5. Check if the result multiplied by the sign is out of the 32-bit signed integer range. If so, return the appropriate boundary value.
6. When no more digits are found, return the final result multiplied by the sign. | LeetCode 8: String to Integer (atoi)
string-to-integer-atoi
Implement the `myAtoi(string s)` function, which converts a string to a 32-bit signed integer (similar to C/C++'s `atoi` function).
The algorithm for `myAtoi(string s)` is as follows:
1. Read in and ignore any leading whitespace.
2. Check if the next character (if not already at the end of the string) is `'-'` or `'+'`. Read this character in if it is either. This determines if the final result is negative or positive respectively. Assume the result is positive if neither is present.
3. Read in next the characters until the next non-digit character or the end of the input is reached. The rest of the string is ignored.
4. Convert these digits into an integer (i.e. `"123 " -> 123`, `"0032 " -> 32`). If no digits were read, then the integer is `0`. Change the sign as necessary (from step 2).
5. If the integer is out of the 32-bit signed integer range `[-231, 231 - 1]`, then clamp the integer so that it remains in the range. Specifically, integers less than `-231` should be clamped to `-231`, and integers greater than `231 - 1` should be clamped to `231 - 1`.
6. Return the integer as the final result.
**Note:**
* Only the space character `' '` is considered a whitespace character.
* **Do not ignore** any characters other than the leading whitespace or the rest of the string after the digits.
**Example 1:**
**Input:** s = "42 "
**Output:** 42
**Explanation:** The underlined characters are what is read in, the caret is the current reader position.
Step 1: "42 " (no characters read because there is no leading whitespace)
^
Step 2: "42 " (no characters read because there is neither a '-' nor '+')
^
Step 3: "42 " ( "42 " is read in)
^
The parsed integer is 42.
Since 42 is in the range \[-231, 231 - 1\], the final result is 42.
**Example 2:**
**Input:** s = " -42 "
**Output:** -42
**Explanation:**
Step 1: " \-42 " (leading whitespace is read and ignored)
^
Step 2: " \-42 " ('-' is read, so the result should be negative)
^
Step 3: " -42 " ( "42 " is read in)
^
The parsed integer is -42.
Since -42 is in the range \[-231, 231 - 1\], the final result is -42.
**Example 3:**
**Input:** s = "4193 with words "
**Output:** 4193
**Explanation:**
Step 1: "4193 with words " (no characters read because there is no leading whitespace)
^
Step 2: "4193 with words " (no characters read because there is neither a '-' nor '+')
^
Step 3: "4193 with words " ( "4193 " is read in; reading stops because the next character is a non-digit)
^
The parsed integer is 4193.
Since 4193 is in the range \[-231, 231 - 1\], the final result is 4193.
**Constraints:**
* `0 <= s.length <= 200`
* `s` consists of English letters (lower-case and upper-case), digits (`0-9`), `' '`, `'+'`, `'-'`, and `'.'`. | Solve the following problem in python, and explain your solution. |
```cpp
bool isPalindrome(int x) {
if (x < 0) return false;
int original = x, reversed = 0;
while (x > 0) {
reversed = reversed * 10 + x % 10;
x /= 10;
}
return original == reversed;
}
```
Explanation:
Here is the algorithm:
1. If the integer is negative, immediately return false, as negative numbers cannot be palindromes.
2. Create a variable named `original` to store the original integer value and `reversed` initialized with 0 to store the reversed integer value.
3. Iterate through the integer while it's greater than zero and, in each iteration, perform the following operations:
a. Multiply the `reversed` value by 10 and add the modulus of the integer (x) by 10.
b. Divide the integer (x) by 10, discarding the remainder.
4. Compare `original` and `reversed` integers; if they are equal, return true; otherwise, return false.
This approach requires O(log10(n)) time complexity as we are dividing the input by 10 in each iteration. | LeetCode 9: Palindrome Number
palindrome-number
Given an integer `x`, return `true` _if_ `x` _is a_ _**palindrome**__, and_ `false` _otherwise_.
**Example 1:**
**Input:** x = 121
**Output:** true
**Explanation:** 121 reads as 121 from left to right and from right to left.
**Example 2:**
**Input:** x = -121
**Output:** false
**Explanation:** From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.
**Example 3:**
**Input:** x = 10
**Output:** false
**Explanation:** Reads 01 from right to left. Therefore it is not a palindrome.
**Constraints:**
* `-231 <= x <= 231 - 1`
**Follow up:** Could you solve it without converting the integer to a string? | Solve the following problem in c++, and explain your solution. |
```java
public boolean isPalindrome(int x) {
if (x < 0) return false;
int original = x, reversed = 0;
while (x > 0) {
reversed = reversed * 10 + x % 10;
x /= 10;
}
return original == reversed;
}
```
Explanation:
Here is the algorithm:
1. If the integer is negative, immediately return false, as negative numbers cannot be palindromes.
2. Create a variable named `original` to store the original integer value and `reversed` initialized with 0 to store the reversed integer value.
3. Iterate through the integer while it's greater than zero and, in each iteration, perform the following operations:
a. Multiply the `reversed` value by 10 and add the modulus of the integer (x) by 10.
b. Divide the integer (x) by 10, discarding the remainder.
4. Compare `original` and `reversed` integers; if they are equal, return true; otherwise, return false.
This approach requires O(log10(n)) time complexity as we are dividing the input by 10 in each iteration. | LeetCode 9: Palindrome Number
palindrome-number
Given an integer `x`, return `true` _if_ `x` _is a_ _**palindrome**__, and_ `false` _otherwise_.
**Example 1:**
**Input:** x = 121
**Output:** true
**Explanation:** 121 reads as 121 from left to right and from right to left.
**Example 2:**
**Input:** x = -121
**Output:** false
**Explanation:** From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.
**Example 3:**
**Input:** x = 10
**Output:** false
**Explanation:** Reads 01 from right to left. Therefore it is not a palindrome.
**Constraints:**
* `-231 <= x <= 231 - 1`
**Follow up:** Could you solve it without converting the integer to a string? | Solve the following problem in java, and explain your solution. |
```javascript
function isPalindrome(x) {
if (x < 0) return false;
let original = x, reversed = 0;
while (x > 0) {
reversed = reversed * 10 + x % 10;
x = Math.floor(x / 10);
}
return original === reversed;
}
```
Explanation:
Here is the algorithm:
1. If the integer is negative, immediately return false, as negative numbers cannot be palindromes.
2. Create a variable named `original` to store the original integer value and `reversed` initialized with 0 to store the reversed integer value.
3. Iterate through the integer while it's greater than zero and, in each iteration, perform the following operations:
a. Multiply the `reversed` value by 10 and add the modulus of the integer (x) by 10.
b. Divide the integer (x) by 10, discarding the remainder.
4. Compare `original` and `reversed` integers; if they are equal, return true; otherwise, return false.
This approach requires O(log10(n)) time complexity as we are dividing the input by 10 in each iteration. | LeetCode 9: Palindrome Number
palindrome-number
Given an integer `x`, return `true` _if_ `x` _is a_ _**palindrome**__, and_ `false` _otherwise_.
**Example 1:**
**Input:** x = 121
**Output:** true
**Explanation:** 121 reads as 121 from left to right and from right to left.
**Example 2:**
**Input:** x = -121
**Output:** false
**Explanation:** From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.
**Example 3:**
**Input:** x = 10
**Output:** false
**Explanation:** Reads 01 from right to left. Therefore it is not a palindrome.
**Constraints:**
* `-231 <= x <= 231 - 1`
**Follow up:** Could you solve it without converting the integer to a string? | Solve the following problem in javascript, and explain your solution. |
```python
def is_palindrome(x: int) -> bool:
if x < 0:
return False
original, reversed = x, 0
while x > 0:
reversed = reversed * 10 + x % 10
x //= 10
return original == reversed
```
Explanation:
Here is the algorithm:
1. If the integer is negative, immediately return false, as negative numbers cannot be palindromes.
2. Create a variable named `original` to store the original integer value and `reversed` initialized with 0 to store the reversed integer value.
3. Iterate through the integer while it's greater than zero and, in each iteration, perform the following operations:
a. Multiply the `reversed` value by 10 and add the modulus of the integer (x) by 10.
b. Divide the integer (x) by 10, discarding the remainder.
4. Compare `original` and `reversed` integers; if they are equal, return true; otherwise, return false.
This approach requires O(log10(n)) time complexity as we are dividing the input by 10 in each iteration. | LeetCode 9: Palindrome Number
palindrome-number
Given an integer `x`, return `true` _if_ `x` _is a_ _**palindrome**__, and_ `false` _otherwise_.
**Example 1:**
**Input:** x = 121
**Output:** true
**Explanation:** 121 reads as 121 from left to right and from right to left.
**Example 2:**
**Input:** x = -121
**Output:** false
**Explanation:** From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.
**Example 3:**
**Input:** x = 10
**Output:** false
**Explanation:** Reads 01 from right to left. Therefore it is not a palindrome.
**Constraints:**
* `-231 <= x <= 231 - 1`
**Follow up:** Could you solve it without converting the integer to a string? | Solve the following problem in python, and explain your solution. |
```cpp
bool isMatch(std::string s, std::string p) {
int m = s.length(), n = p.length();
std::vector<std::vector<bool>> dp(m + 1, std::vector<bool>(n + 1, false));
dp[0][0] = true;
for (int j = 1; j <= n; j++) {
if (p[j - 1] == '*' && dp[0][j - 2]) {
dp[0][j] = true;
}
}
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
if (p[j - 1] == s[i - 1] || p[j - 1] == '.') {
dp[i][j] = dp[i - 1][j - 1];
} else if (p[j - 1] == '*') {
dp[i][j] = dp[i][j - 2] || (dp[i - 1][j] && (s[i - 1] == p[j - 2] || p[j - 2] == '.'));
}
}
}
return dp[m][n];
}
```
Explanation:
We use Dynamic Programming to solve this problem. We store intermediate results in a 2D boolean array `dp`, where `dp[i][j]` indicates whether the first `i` characters of `s` match the first `j` characters of `p`. We then loop through the characters of both strings and update the `dp` table.
If the current characters in `s` and `p` are equal or `p` has a '.', we update the `dp[i][j]` to the value `dp[i - 1][j - 1]`, because the match continues.
If the current character in `p` is '*', we have two options:
1. Consider the matching of the previous character in `p` as 0 occurrences. We update the `dp[i][j]` value to `dp[i][j - 2]`.
2. Consider the matching of the previous character in `p` as at least 1 occurrence. We update `dp[i][j]` to the value of `dp[i - 1][j] && (s[i - 1] == p[j - 2] || p[j - 2] == '.')`.
Finally, `dp[m][n]` stores the result, where `m` and `n` are the lengths of strings `s` and `p` respectively. | LeetCode 10: Regular Expression Matching
regular-expression-matching
Given an input string `s` and a pattern `p`, implement regular expression matching with support for `'.'` and `'*'` where:
* `'.'` Matches any single character.
* `'*'` Matches zero or more of the preceding element.
The matching should cover the **entire** input string (not partial).
**Example 1:**
**Input:** s = "aa ", p = "a "
**Output:** false
**Explanation:** "a " does not match the entire string "aa ".
**Example 2:**
**Input:** s = "aa ", p = "a\* "
**Output:** true
**Explanation:** '\*' means zero or more of the preceding element, 'a'. Therefore, by repeating 'a' once, it becomes "aa ".
**Example 3:**
**Input:** s = "ab ", p = ".\* "
**Output:** true
**Explanation:** ".\* " means "zero or more (\*) of any character (.) ".
**Constraints:**
* `1 <= s.length <= 20`
* `1 <= p.length <= 20`
* `s` contains only lowercase English letters.
* `p` contains only lowercase English letters, `'.'`, and `'*'`.
* It is guaranteed for each appearance of the character `'*'`, there will be a previous valid character to match. | Solve the following problem in c++, and explain your solution. |
```java
public boolean isMatch(String s, String p) {
int m = s.length(), n = p.length();
boolean[][] dp = new boolean[m + 1][n + 1];
dp[0][0] = true;
for (int j = 1; j <= n; j++) {
if (p.charAt(j - 1) == '*' && dp[0][j - 2]) {
dp[0][j] = true;
}
}
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
if (p.charAt(j - 1) == s.charAt(i - 1) || p.charAt(j - 1) == '.') {
dp[i][j] = dp[i - 1][j - 1];
} else if (p.charAt(j - 1) == '*') {
dp[i][j] = dp[i][j - 2] || (dp[i - 1][j] && (s.charAt(i - 1) == p.charAt(j - 2) || p.charAt(j - 2) == '.'));
}
}
}
return dp[m][n];
}
```
Explanation:
We use Dynamic Programming to solve this problem. We store intermediate results in a 2D boolean array `dp`, where `dp[i][j]` indicates whether the first `i` characters of `s` match the first `j` characters of `p`. We then loop through the characters of both strings and update the `dp` table.
If the current characters in `s` and `p` are equal or `p` has a '.', we update the `dp[i][j]` to the value `dp[i - 1][j - 1]`, because the match continues.
If the current character in `p` is '*', we have two options:
1. Consider the matching of the previous character in `p` as 0 occurrences. We update the `dp[i][j]` value to `dp[i][j - 2]`.
2. Consider the matching of the previous character in `p` as at least 1 occurrence. We update `dp[i][j]` to the value of `dp[i - 1][j] && (s[i - 1] == p[j - 2] || p[j - 2] == '.')`.
Finally, `dp[m][n]` stores the result, where `m` and `n` are the lengths of strings `s` and `p` respectively. | LeetCode 10: Regular Expression Matching
regular-expression-matching
Given an input string `s` and a pattern `p`, implement regular expression matching with support for `'.'` and `'*'` where:
* `'.'` Matches any single character.
* `'*'` Matches zero or more of the preceding element.
The matching should cover the **entire** input string (not partial).
**Example 1:**
**Input:** s = "aa ", p = "a "
**Output:** false
**Explanation:** "a " does not match the entire string "aa ".
**Example 2:**
**Input:** s = "aa ", p = "a\* "
**Output:** true
**Explanation:** '\*' means zero or more of the preceding element, 'a'. Therefore, by repeating 'a' once, it becomes "aa ".
**Example 3:**
**Input:** s = "ab ", p = ".\* "
**Output:** true
**Explanation:** ".\* " means "zero or more (\*) of any character (.) ".
**Constraints:**
* `1 <= s.length <= 20`
* `1 <= p.length <= 20`
* `s` contains only lowercase English letters.
* `p` contains only lowercase English letters, `'.'`, and `'*'`.
* It is guaranteed for each appearance of the character `'*'`, there will be a previous valid character to match. | Solve the following problem in java, and explain your solution. |
```javascript
function isMatch(s, p) {
const m = s.length, n = p.length;
const dp = Array.from({ length: m + 1 }, () => Array(n + 1).fill(false));
dp[0][0] = true;
for (let j = 1; j <= n; j++) {
if (p[j - 1] === '*' && dp[0][j - 2]) {
dp[0][j] = true;
}
}
for (let i = 1; i <= m; i++) {
for (let j = 1; j <= n; j++) {
if (p[j - 1] === s[i - 1] || p[j - 1] === '.') {
dp[i][j] = dp[i - 1][j - 1];
} else if (p[j - 1] === '*') {
dp[i][j] = dp[i][j - 2] || (dp[i - 1][j] && (s[i - 1] === p[j - 2] || p[j - 2] === '.'));
}
}
}
return dp[m][n];
}
```
Explanation:
We use Dynamic Programming to solve this problem. We store intermediate results in a 2D boolean array `dp`, where `dp[i][j]` indicates whether the first `i` characters of `s` match the first `j` characters of `p`. We then loop through the characters of both strings and update the `dp` table.
If the current characters in `s` and `p` are equal or `p` has a '.', we update the `dp[i][j]` to the value `dp[i - 1][j - 1]`, because the match continues.
If the current character in `p` is '*', we have two options:
1. Consider the matching of the previous character in `p` as 0 occurrences. We update the `dp[i][j]` value to `dp[i][j - 2]`.
2. Consider the matching of the previous character in `p` as at least 1 occurrence. We update `dp[i][j]` to the value of `dp[i - 1][j] && (s[i - 1] == p[j - 2] || p[j - 2] == '.')`.
Finally, `dp[m][n]` stores the result, where `m` and `n` are the lengths of strings `s` and `p` respectively. | LeetCode 10: Regular Expression Matching
regular-expression-matching
Given an input string `s` and a pattern `p`, implement regular expression matching with support for `'.'` and `'*'` where:
* `'.'` Matches any single character.
* `'*'` Matches zero or more of the preceding element.
The matching should cover the **entire** input string (not partial).
**Example 1:**
**Input:** s = "aa ", p = "a "
**Output:** false
**Explanation:** "a " does not match the entire string "aa ".
**Example 2:**
**Input:** s = "aa ", p = "a\* "
**Output:** true
**Explanation:** '\*' means zero or more of the preceding element, 'a'. Therefore, by repeating 'a' once, it becomes "aa ".
**Example 3:**
**Input:** s = "ab ", p = ".\* "
**Output:** true
**Explanation:** ".\* " means "zero or more (\*) of any character (.) ".
**Constraints:**
* `1 <= s.length <= 20`
* `1 <= p.length <= 20`
* `s` contains only lowercase English letters.
* `p` contains only lowercase English letters, `'.'`, and `'*'`.
* It is guaranteed for each appearance of the character `'*'`, there will be a previous valid character to match. | Solve the following problem in javascript, and explain your solution. |
```python
def is_match(s: str, p: str) -> bool:
m, n = len(s), len(p)
dp = [[False] * (n + 1) for _ in range(m + 1)]
dp[0][0] = True
for j in range(1, n + 1):
if p[j - 1] == '*' and dp[0][j - 2]:
dp[0][j] = True
for i in range(1, m + 1):
for j in range(1, n + 1):
if p[j - 1] == s[i - 1] or p[j - 1] == '.':
dp[i][j] = dp[i - 1][j - 1]
elif p[j - 1] == '*':
dp[i][j] = dp[i][j - 2] or (dp[i - 1][j] and (s[i - 1] == p[j - 2] or p[j - 2] == '.'))
return dp[m][n]
```
Explanation:
We use Dynamic Programming to solve this problem. We store intermediate results in a 2D boolean array `dp`, where `dp[i][j]` indicates whether the first `i` characters of `s` match the first `j` characters of `p`. We then loop through the characters of both strings and update the `dp` table.
If the current characters in `s` and `p` are equal or `p` has a '.', we update the `dp[i][j]` to the value `dp[i - 1][j - 1]`, because the match continues.
If the current character in `p` is '*', we have two options:
1. Consider the matching of the previous character in `p` as 0 occurrences. We update the `dp[i][j]` value to `dp[i][j - 2]`.
2. Consider the matching of the previous character in `p` as at least 1 occurrence. We update `dp[i][j]` to the value of `dp[i - 1][j] && (s[i - 1] == p[j - 2] || p[j - 2] == '.')`.
Finally, `dp[m][n]` stores the result, where `m` and `n` are the lengths of strings `s` and `p` respectively. | LeetCode 10: Regular Expression Matching
regular-expression-matching
Given an input string `s` and a pattern `p`, implement regular expression matching with support for `'.'` and `'*'` where:
* `'.'` Matches any single character.
* `'*'` Matches zero or more of the preceding element.
The matching should cover the **entire** input string (not partial).
**Example 1:**
**Input:** s = "aa ", p = "a "
**Output:** false
**Explanation:** "a " does not match the entire string "aa ".
**Example 2:**
**Input:** s = "aa ", p = "a\* "
**Output:** true
**Explanation:** '\*' means zero or more of the preceding element, 'a'. Therefore, by repeating 'a' once, it becomes "aa ".
**Example 3:**
**Input:** s = "ab ", p = ".\* "
**Output:** true
**Explanation:** ".\* " means "zero or more (\*) of any character (.) ".
**Constraints:**
* `1 <= s.length <= 20`
* `1 <= p.length <= 20`
* `s` contains only lowercase English letters.
* `p` contains only lowercase English letters, `'.'`, and `'*'`.
* It is guaranteed for each appearance of the character `'*'`, there will be a previous valid character to match. | Solve the following problem in python, and explain your solution. |
```cpp
int maxArea(vector<int>& height) {
int max_area = 0, left = 0, right = height.size() - 1;
while (left < right) {
max_area = max(max_area, min(height[left], height[right]) * (right - left));
if (height[left] < height[right]) {
left++;
} else {
right--;
}
}
return max_area;
}
```
Explanation:
The algorithm uses a two-pointer approach, with one pointer starting from the left end and the other from the right end. It calculates the area between these two lines and updates the maximum area if the current area is larger than the previously calculated maximum area. If the height at the left pointer is less than the height at the right pointer, it moves the left pointer to the right. Otherwise, it moves the right pointer to the left. The algorithm continues until the pointers meet.
This algorithm ensures that the maximum area is found without having to check all possible pairs of lines because it always chooses the pair of lines with the maximum possible distance and minimum height difference at each step. | LeetCode 11: Container With Most Water
container-with-most-water
You are given an integer array `height` of length `n`. There are `n` vertical lines drawn such that the two endpoints of the `ith` line are `(i, 0)` and `(i, height[i])`.
Find two lines that together with the x-axis form a container, such that the container contains the most water.
Return _the maximum amount of water a container can store_.
**Notice** that you may not slant the container.
**Example 1:**
**Input:** height = \[1,8,6,2,5,4,8,3,7\]
**Output:** 49
**Explanation:** The above vertical lines are represented by array \[1,8,6,2,5,4,8,3,7\]. In this case, the max area of water (blue section) the container can contain is 49.
**Example 2:**
**Input:** height = \[1,1\]
**Output:** 1
**Constraints:**
* `n == height.length`
* `2 <= n <= 105`
* `0 <= height[i] <= 104` | Solve the following problem in c++, and explain your solution. |
```java
public int maxArea(int[] height) {
int max_area = 0, left = 0, right = height.length - 1;
while (left < right) {
max_area = Math.max(max_area, Math.min(height[left], height[right]) * (right - left));
if (height[left] < height[right]) {
left++;
} else {
right--;
}
}
return max_area;
}
```
Explanation:
The algorithm uses a two-pointer approach, with one pointer starting from the left end and the other from the right end. It calculates the area between these two lines and updates the maximum area if the current area is larger than the previously calculated maximum area. If the height at the left pointer is less than the height at the right pointer, it moves the left pointer to the right. Otherwise, it moves the right pointer to the left. The algorithm continues until the pointers meet.
This algorithm ensures that the maximum area is found without having to check all possible pairs of lines because it always chooses the pair of lines with the maximum possible distance and minimum height difference at each step. | LeetCode 11: Container With Most Water
container-with-most-water
You are given an integer array `height` of length `n`. There are `n` vertical lines drawn such that the two endpoints of the `ith` line are `(i, 0)` and `(i, height[i])`.
Find two lines that together with the x-axis form a container, such that the container contains the most water.
Return _the maximum amount of water a container can store_.
**Notice** that you may not slant the container.
**Example 1:**
**Input:** height = \[1,8,6,2,5,4,8,3,7\]
**Output:** 49
**Explanation:** The above vertical lines are represented by array \[1,8,6,2,5,4,8,3,7\]. In this case, the max area of water (blue section) the container can contain is 49.
**Example 2:**
**Input:** height = \[1,1\]
**Output:** 1
**Constraints:**
* `n == height.length`
* `2 <= n <= 105`
* `0 <= height[i] <= 104` | Solve the following problem in java, and explain your solution. |
```javascript
function maxArea(height) {
let max_area = 0, left = 0, right = height.length - 1;
while (left < right) {
max_area = Math.max(max_area, Math.min(height[left], height[right]) * (right - left));
if (height[left] < height[right]) {
left++;
} else {
right--;
}
}
return max_area;
}
```
Explanation:
The algorithm uses a two-pointer approach, with one pointer starting from the left end and the other from the right end. It calculates the area between these two lines and updates the maximum area if the current area is larger than the previously calculated maximum area. If the height at the left pointer is less than the height at the right pointer, it moves the left pointer to the right. Otherwise, it moves the right pointer to the left. The algorithm continues until the pointers meet.
This algorithm ensures that the maximum area is found without having to check all possible pairs of lines because it always chooses the pair of lines with the maximum possible distance and minimum height difference at each step. | LeetCode 11: Container With Most Water
container-with-most-water
You are given an integer array `height` of length `n`. There are `n` vertical lines drawn such that the two endpoints of the `ith` line are `(i, 0)` and `(i, height[i])`.
Find two lines that together with the x-axis form a container, such that the container contains the most water.
Return _the maximum amount of water a container can store_.
**Notice** that you may not slant the container.
**Example 1:**
**Input:** height = \[1,8,6,2,5,4,8,3,7\]
**Output:** 49
**Explanation:** The above vertical lines are represented by array \[1,8,6,2,5,4,8,3,7\]. In this case, the max area of water (blue section) the container can contain is 49.
**Example 2:**
**Input:** height = \[1,1\]
**Output:** 1
**Constraints:**
* `n == height.length`
* `2 <= n <= 105`
* `0 <= height[i] <= 104` | Solve the following problem in javascript, and explain your solution. |
```python
def max_area(height):
max_area, left, right = 0, 0, len(height) - 1
while left < right:
max_area = max(max_area, min(height[left], height[right]) * (right - left))
if height[left] < height[right]:
left += 1
else:
right -= 1
return max_area
```
Explanation:
The algorithm uses a two-pointer approach, with one pointer starting from the left end and the other from the right end. It calculates the area between these two lines and updates the maximum area if the current area is larger than the previously calculated maximum area. If the height at the left pointer is less than the height at the right pointer, it moves the left pointer to the right. Otherwise, it moves the right pointer to the left. The algorithm continues until the pointers meet.
This algorithm ensures that the maximum area is found without having to check all possible pairs of lines because it always chooses the pair of lines with the maximum possible distance and minimum height difference at each step. | LeetCode 11: Container With Most Water
container-with-most-water
You are given an integer array `height` of length `n`. There are `n` vertical lines drawn such that the two endpoints of the `ith` line are `(i, 0)` and `(i, height[i])`.
Find two lines that together with the x-axis form a container, such that the container contains the most water.
Return _the maximum amount of water a container can store_.
**Notice** that you may not slant the container.
**Example 1:**
**Input:** height = \[1,8,6,2,5,4,8,3,7\]
**Output:** 49
**Explanation:** The above vertical lines are represented by array \[1,8,6,2,5,4,8,3,7\]. In this case, the max area of water (blue section) the container can contain is 49.
**Example 2:**
**Input:** height = \[1,1\]
**Output:** 1
**Constraints:**
* `n == height.length`
* `2 <= n <= 105`
* `0 <= height[i] <= 104` | Solve the following problem in python, and explain your solution. |
```cpp
#include <vector>
#include <string>
std::string intToRoman(int num) {
std::vector<std::pair<int, std::string>> romans = {
{1000, "M"},
{900, "CM"},
{500, "D"},
{400, "CD"},
{100, "C"},
{90, "XC"},
{50, "L"},
{40, "XL"},
{10, "X"},
{9, "IX"},
{5, "V"},
{4, "IV"},
{1, "I"}
};
std::string roman = "";
for (const auto& p : romans) {
while (num >= p.first) {
roman += p.second;
num -= p.first;
}
}
return roman;
}
```
Explanation:
The algorithm starts with a list of pairs, where each pair contains a decimal value and its corresponding roman numeral. Then, it iteratively go through the list of pairs and checks if the num is greater than or equal to the decimal value. If it is, it appends the roman numeral to the result, and subtracts the decimal value from the num. This is done until the num becomes 0. | LeetCode 12: Integer to Roman
integer-to-roman
Roman numerals are represented by seven different symbols: `I`, `V`, `X`, `L`, `C`, `D` and `M`.
**Symbol** **Value**
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
For example, `2` is written as `II` in Roman numeral, just two one's added together. `12` is written as `XII`, which is simply `X + II`. The number `27` is written as `XXVII`, which is `XX + V + II`.
Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not `IIII`. Instead, the number four is written as `IV`. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as `IX`. There are six instances where subtraction is used:
* `I` can be placed before `V` (5) and `X` (10) to make 4 and 9.
* `X` can be placed before `L` (50) and `C` (100) to make 40 and 90.
* `C` can be placed before `D` (500) and `M` (1000) to make 400 and 900.
Given an integer, convert it to a roman numeral.
**Example 1:**
**Input:** num = 3
**Output:** "III "
**Explanation:** 3 is represented as 3 ones.
**Example 2:**
**Input:** num = 58
**Output:** "LVIII "
**Explanation:** L = 50, V = 5, III = 3.
**Example 3:**
**Input:** num = 1994
**Output:** "MCMXCIV "
**Explanation:** M = 1000, CM = 900, XC = 90 and IV = 4.
**Constraints:**
* `1 <= num <= 3999` | Solve the following problem in c++, and explain your solution. |
```java
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
public class RomanNumerals {
public static String intToRoman(int num) {
List<Pair> romans = Arrays.asList(
new Pair(1000, "M"), new Pair(900, "CM"), new Pair(500, "D"),
new Pair(400, "CD"), new Pair(100, "C"), new Pair(90, "XC"),
new Pair(50, "L"), new Pair(40, "XL"), new Pair(10, "X"),
new Pair(9, "IX"), new Pair(5, "V"), new Pair(4, "IV"),
new Pair(1, "I")
);
StringBuilder roman = new StringBuilder();
for (Pair p : romans) {
while (num >= p.num) {
roman.append(p.symbol);
num -= p.num;
}
}
return roman.toString();
}
private static class Pair {
final int num;
final String symbol;
Pair(int num, String symbol) {
this.num = num;
this.symbol = symbol;
}
}
}
```
Explanation:
The algorithm starts with a list of pairs, where each pair contains a decimal value and its corresponding roman numeral. Then, it iteratively go through the list of pairs and checks if the num is greater than or equal to the decimal value. If it is, it appends the roman numeral to the result, and subtracts the decimal value from the num. This is done until the num becomes 0. | LeetCode 12: Integer to Roman
integer-to-roman
Roman numerals are represented by seven different symbols: `I`, `V`, `X`, `L`, `C`, `D` and `M`.
**Symbol** **Value**
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
For example, `2` is written as `II` in Roman numeral, just two one's added together. `12` is written as `XII`, which is simply `X + II`. The number `27` is written as `XXVII`, which is `XX + V + II`.
Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not `IIII`. Instead, the number four is written as `IV`. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as `IX`. There are six instances where subtraction is used:
* `I` can be placed before `V` (5) and `X` (10) to make 4 and 9.
* `X` can be placed before `L` (50) and `C` (100) to make 40 and 90.
* `C` can be placed before `D` (500) and `M` (1000) to make 400 and 900.
Given an integer, convert it to a roman numeral.
**Example 1:**
**Input:** num = 3
**Output:** "III "
**Explanation:** 3 is represented as 3 ones.
**Example 2:**
**Input:** num = 58
**Output:** "LVIII "
**Explanation:** L = 50, V = 5, III = 3.
**Example 3:**
**Input:** num = 1994
**Output:** "MCMXCIV "
**Explanation:** M = 1000, CM = 900, XC = 90 and IV = 4.
**Constraints:**
* `1 <= num <= 3999` | Solve the following problem in java, and explain your solution. |
```javascript
function intToRoman(num) {
const romans = [
[1000, "M"], [900, "CM"], [500, "D"],
[400, "CD"], [100, "C"], [90, "XC"],
[50, "L"], [40, "XL"], [10, "X"],
[9, "IX"], [5, "V"], [4, "IV"], [1, "I"]
];
let roman = "";
for (const [value, symbol] of romans) {
while (num >= value) {
roman += symbol;
num -= value;
}
}
return roman;
}
```
Explanation:
The algorithm starts with a list of pairs, where each pair contains a decimal value and its corresponding roman numeral. Then, it iteratively go through the list of pairs and checks if the num is greater than or equal to the decimal value. If it is, it appends the roman numeral to the result, and subtracts the decimal value from the num. This is done until the num becomes 0. | LeetCode 12: Integer to Roman
integer-to-roman
Roman numerals are represented by seven different symbols: `I`, `V`, `X`, `L`, `C`, `D` and `M`.
**Symbol** **Value**
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
For example, `2` is written as `II` in Roman numeral, just two one's added together. `12` is written as `XII`, which is simply `X + II`. The number `27` is written as `XXVII`, which is `XX + V + II`.
Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not `IIII`. Instead, the number four is written as `IV`. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as `IX`. There are six instances where subtraction is used:
* `I` can be placed before `V` (5) and `X` (10) to make 4 and 9.
* `X` can be placed before `L` (50) and `C` (100) to make 40 and 90.
* `C` can be placed before `D` (500) and `M` (1000) to make 400 and 900.
Given an integer, convert it to a roman numeral.
**Example 1:**
**Input:** num = 3
**Output:** "III "
**Explanation:** 3 is represented as 3 ones.
**Example 2:**
**Input:** num = 58
**Output:** "LVIII "
**Explanation:** L = 50, V = 5, III = 3.
**Example 3:**
**Input:** num = 1994
**Output:** "MCMXCIV "
**Explanation:** M = 1000, CM = 900, XC = 90 and IV = 4.
**Constraints:**
* `1 <= num <= 3999` | Solve the following problem in javascript, and explain your solution. |
```python
def int_to_roman(num):
romans = [
(1000, "M"), (900, "CM"), (500, "D"),
(400, "CD"), (100, "C"), (90, "XC"),
(50, "L"), (40, "XL"), (10, "X"),
(9, "IX"), (5, "V"), (4, "IV"), (1, "I")
]
roman = ""
for value, symbol in romans:
while num >= value:
roman += symbol
num -= value
return roman
```
Explanation:
The algorithm starts with a list of pairs, where each pair contains a decimal value and its corresponding roman numeral. Then, it iteratively go through the list of pairs and checks if the num is greater than or equal to the decimal value. If it is, it appends the roman numeral to the result, and subtracts the decimal value from the num. This is done until the num becomes 0. | LeetCode 12: Integer to Roman
integer-to-roman
Roman numerals are represented by seven different symbols: `I`, `V`, `X`, `L`, `C`, `D` and `M`.
**Symbol** **Value**
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
For example, `2` is written as `II` in Roman numeral, just two one's added together. `12` is written as `XII`, which is simply `X + II`. The number `27` is written as `XXVII`, which is `XX + V + II`.
Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not `IIII`. Instead, the number four is written as `IV`. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as `IX`. There are six instances where subtraction is used:
* `I` can be placed before `V` (5) and `X` (10) to make 4 and 9.
* `X` can be placed before `L` (50) and `C` (100) to make 40 and 90.
* `C` can be placed before `D` (500) and `M` (1000) to make 400 and 900.
Given an integer, convert it to a roman numeral.
**Example 1:**
**Input:** num = 3
**Output:** "III "
**Explanation:** 3 is represented as 3 ones.
**Example 2:**
**Input:** num = 58
**Output:** "LVIII "
**Explanation:** L = 50, V = 5, III = 3.
**Example 3:**
**Input:** num = 1994
**Output:** "MCMXCIV "
**Explanation:** M = 1000, CM = 900, XC = 90 and IV = 4.
**Constraints:**
* `1 <= num <= 3999` | Solve the following problem in python, and explain your solution. |
```cpp
int romanToInt(string s) {
unordered_map<char, int> roman_values = { {'I', 1}, {'V', 5}, {'X', 10}, {'L', 50}, {'C', 100}, {'D', 500}, {'M', 1000} };
int total = 0;
int prev_value = 0;
for (char c : s) {
int current_value = roman_values[c];
if (current_value > prev_value)
total += current_value - (2 * prev_value);
else
total += current_value;
prev_value = current_value;
}
return total;
}
```
Explanation:
The algorithm iterates through the input string, which represents the Roman numeral. At each iteration, the character is converted to its corresponding integer value using a hash map. Then, the value is added to the total. However, if the current value is greater than the previous value, it means that we need to subtract the previous value from the current value (while considering that the previous value has already been added once). To do this, we add to the total the current value minus 2 times the previous value. Finally, the previous value is updated, and the loop continues. After the loop finishes, the total gives us the integer value equivalent to the Roman numeral. | LeetCode 13: Roman to Integer
roman-to-integer
Roman numerals are represented by seven different symbols: `I`, `V`, `X`, `L`, `C`, `D` and `M`.
**Symbol** **Value**
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
For example, `2` is written as `II` in Roman numeral, just two ones added together. `12` is written as `XII`, which is simply `X + II`. The number `27` is written as `XXVII`, which is `XX + V + II`.
Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not `IIII`. Instead, the number four is written as `IV`. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as `IX`. There are six instances where subtraction is used:
* `I` can be placed before `V` (5) and `X` (10) to make 4 and 9.
* `X` can be placed before `L` (50) and `C` (100) to make 40 and 90.
* `C` can be placed before `D` (500) and `M` (1000) to make 400 and 900.
Given a roman numeral, convert it to an integer.
**Example 1:**
**Input:** s = "III "
**Output:** 3
**Explanation:** III = 3.
**Example 2:**
**Input:** s = "LVIII "
**Output:** 58
**Explanation:** L = 50, V= 5, III = 3.
**Example 3:**
**Input:** s = "MCMXCIV "
**Output:** 1994
**Explanation:** M = 1000, CM = 900, XC = 90 and IV = 4.
**Constraints:**
* `1 <= s.length <= 15`
* `s` contains only the characters `('I', 'V', 'X', 'L', 'C', 'D', 'M')`.
* It is **guaranteed** that `s` is a valid roman numeral in the range `[1, 3999]`. | Solve the following problem in c++, and explain your solution. |
```java
public int romanToInt(String s) {
Map<Character, Integer> romanValues = new HashMap<>();
romanValues.put('I', 1);
romanValues.put('V', 5);
romanValues.put('X', 10);
romanValues.put('L', 50);
romanValues.put('C', 100);
romanValues.put('D', 500);
romanValues.put('M', 1000);
int total = 0;
int prevValue = 0;
for (char c : s.toCharArray()) {
int currValue = romanValues.get(c);
total += currValue > prevValue ? currValue - 2 * prevValue : currValue;
prevValue = currValue;
}
return total;
}
```
Explanation:
The algorithm iterates through the input string, which represents the Roman numeral. At each iteration, the character is converted to its corresponding integer value using a hash map. Then, the value is added to the total. However, if the current value is greater than the previous value, it means that we need to subtract the previous value from the current value (while considering that the previous value has already been added once). To do this, we add to the total the current value minus 2 times the previous value. Finally, the previous value is updated, and the loop continues. After the loop finishes, the total gives us the integer value equivalent to the Roman numeral. | LeetCode 13: Roman to Integer
roman-to-integer
Roman numerals are represented by seven different symbols: `I`, `V`, `X`, `L`, `C`, `D` and `M`.
**Symbol** **Value**
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
For example, `2` is written as `II` in Roman numeral, just two ones added together. `12` is written as `XII`, which is simply `X + II`. The number `27` is written as `XXVII`, which is `XX + V + II`.
Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not `IIII`. Instead, the number four is written as `IV`. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as `IX`. There are six instances where subtraction is used:
* `I` can be placed before `V` (5) and `X` (10) to make 4 and 9.
* `X` can be placed before `L` (50) and `C` (100) to make 40 and 90.
* `C` can be placed before `D` (500) and `M` (1000) to make 400 and 900.
Given a roman numeral, convert it to an integer.
**Example 1:**
**Input:** s = "III "
**Output:** 3
**Explanation:** III = 3.
**Example 2:**
**Input:** s = "LVIII "
**Output:** 58
**Explanation:** L = 50, V= 5, III = 3.
**Example 3:**
**Input:** s = "MCMXCIV "
**Output:** 1994
**Explanation:** M = 1000, CM = 900, XC = 90 and IV = 4.
**Constraints:**
* `1 <= s.length <= 15`
* `s` contains only the characters `('I', 'V', 'X', 'L', 'C', 'D', 'M')`.
* It is **guaranteed** that `s` is a valid roman numeral in the range `[1, 3999]`. | Solve the following problem in java, and explain your solution. |
```javascript
function romanToInt(s) {
const romanValues = {'I': 1, 'V': 5, 'X': 10, 'L': 50, 'C': 100, 'D': 500, 'M': 1000};
let total = 0;
let prevValue = 0;
for (let i = 0; i < s.length; i++) {
let currValue = romanValues[s[i]];
if (currValue > prevValue)
total += currValue - (2 * prevValue);
else
total += currValue;
prevValue = currValue;
}
return total;
}
```
Explanation:
The algorithm iterates through the input string, which represents the Roman numeral. At each iteration, the character is converted to its corresponding integer value using a hash map. Then, the value is added to the total. However, if the current value is greater than the previous value, it means that we need to subtract the previous value from the current value (while considering that the previous value has already been added once). To do this, we add to the total the current value minus 2 times the previous value. Finally, the previous value is updated, and the loop continues. After the loop finishes, the total gives us the integer value equivalent to the Roman numeral. | LeetCode 13: Roman to Integer
roman-to-integer
Roman numerals are represented by seven different symbols: `I`, `V`, `X`, `L`, `C`, `D` and `M`.
**Symbol** **Value**
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
For example, `2` is written as `II` in Roman numeral, just two ones added together. `12` is written as `XII`, which is simply `X + II`. The number `27` is written as `XXVII`, which is `XX + V + II`.
Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not `IIII`. Instead, the number four is written as `IV`. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as `IX`. There are six instances where subtraction is used:
* `I` can be placed before `V` (5) and `X` (10) to make 4 and 9.
* `X` can be placed before `L` (50) and `C` (100) to make 40 and 90.
* `C` can be placed before `D` (500) and `M` (1000) to make 400 and 900.
Given a roman numeral, convert it to an integer.
**Example 1:**
**Input:** s = "III "
**Output:** 3
**Explanation:** III = 3.
**Example 2:**
**Input:** s = "LVIII "
**Output:** 58
**Explanation:** L = 50, V= 5, III = 3.
**Example 3:**
**Input:** s = "MCMXCIV "
**Output:** 1994
**Explanation:** M = 1000, CM = 900, XC = 90 and IV = 4.
**Constraints:**
* `1 <= s.length <= 15`
* `s` contains only the characters `('I', 'V', 'X', 'L', 'C', 'D', 'M')`.
* It is **guaranteed** that `s` is a valid roman numeral in the range `[1, 3999]`. | Solve the following problem in javascript, and explain your solution. |
```python
def roman_to_int(s):
roman_values = {'I': 1, 'V': 5, 'X': 10, 'L': 50, 'C': 100, 'D': 500, 'M': 1000}
total = 0
prev_value = 0
for c in s:
current_value = roman_values[c]
if current_value > prev_value:
total += current_value - 2 * prev_value
else:
total += current_value
prev_value = current_value
return total
```
Explanation:
The algorithm iterates through the input string, which represents the Roman numeral. At each iteration, the character is converted to its corresponding integer value using a hash map. Then, the value is added to the total. However, if the current value is greater than the previous value, it means that we need to subtract the previous value from the current value (while considering that the previous value has already been added once). To do this, we add to the total the current value minus 2 times the previous value. Finally, the previous value is updated, and the loop continues. After the loop finishes, the total gives us the integer value equivalent to the Roman numeral. | LeetCode 13: Roman to Integer
roman-to-integer
Roman numerals are represented by seven different symbols: `I`, `V`, `X`, `L`, `C`, `D` and `M`.
**Symbol** **Value**
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
For example, `2` is written as `II` in Roman numeral, just two ones added together. `12` is written as `XII`, which is simply `X + II`. The number `27` is written as `XXVII`, which is `XX + V + II`.
Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not `IIII`. Instead, the number four is written as `IV`. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as `IX`. There are six instances where subtraction is used:
* `I` can be placed before `V` (5) and `X` (10) to make 4 and 9.
* `X` can be placed before `L` (50) and `C` (100) to make 40 and 90.
* `C` can be placed before `D` (500) and `M` (1000) to make 400 and 900.
Given a roman numeral, convert it to an integer.
**Example 1:**
**Input:** s = "III "
**Output:** 3
**Explanation:** III = 3.
**Example 2:**
**Input:** s = "LVIII "
**Output:** 58
**Explanation:** L = 50, V= 5, III = 3.
**Example 3:**
**Input:** s = "MCMXCIV "
**Output:** 1994
**Explanation:** M = 1000, CM = 900, XC = 90 and IV = 4.
**Constraints:**
* `1 <= s.length <= 15`
* `s` contains only the characters `('I', 'V', 'X', 'L', 'C', 'D', 'M')`.
* It is **guaranteed** that `s` is a valid roman numeral in the range `[1, 3999]`. | Solve the following problem in python, and explain your solution. |
```cpp
#include <vector>
#include <string>
using namespace std;
string longestCommonPrefix(vector<string>& strs) {
if (strs.empty()) return "";
for (int i = 0; i < strs[0].size(); ++i) {
char c = strs[0][i];
for (int j = 1; j < strs.size(); ++j) {
if (i == strs[j].size() || strs[j][i] != c) {
return strs[0].substr(0, i);
}
}
}
return strs[0];
}
```
Explanation:
The algorithm iterates through the characters in the first string and compares the equivalent characters in the subsequent strings. If there is a mismatch detected, return the common prefix up to the mismatch point. If the algorithm finishes iterating over the first string, return the entire first string as the common prefix.
1. Check if the input array is empty, if it is, return an empty string as there is no common prefix.
2. Iterate through the characters of the first string.
3. For each character, iterate through the rest of the strings in the array.
4. Compare the current character with the corresponding character in the other strings.
5. If there is a mismatch or if we reach the end of a string, return the common prefix substring up to the current index.
6. If the loop reaches the end of the first string, then the whole first string is the common prefix. Return the entire first string. | LeetCode 14: Longest Common Prefix
longest-common-prefix
Write a function to find the longest common prefix string amongst an array of strings.
If there is no common prefix, return an empty string `" "`.
**Example 1:**
**Input:** strs = \[ "flower ", "flow ", "flight "\]
**Output:** "fl "
**Example 2:**
**Input:** strs = \[ "dog ", "racecar ", "car "\]
**Output:** " "
**Explanation:** There is no common prefix among the input strings.
**Constraints:**
* `1 <= strs.length <= 200`
* `0 <= strs[i].length <= 200`
* `strs[i]` consists of only lowercase English letters. | Solve the following problem in c++, and explain your solution. |
```java
public String longestCommonPrefix(String[] strs) {
if (strs.length == 0) return "";
for (int i = 0; i < strs[0].length(); ++i) {
char c = strs[0].charAt(i);
for (int j = 1; j < strs.length; ++j) {
if (i == strs[j].length() || strs[j].charAt(i) != c) {
return strs[0].substring(0, i);
}
}
}
return strs[0];
}
```
Explanation:
The algorithm iterates through the characters in the first string and compares the equivalent characters in the subsequent strings. If there is a mismatch detected, return the common prefix up to the mismatch point. If the algorithm finishes iterating over the first string, return the entire first string as the common prefix.
1. Check if the input array is empty, if it is, return an empty string as there is no common prefix.
2. Iterate through the characters of the first string.
3. For each character, iterate through the rest of the strings in the array.
4. Compare the current character with the corresponding character in the other strings.
5. If there is a mismatch or if we reach the end of a string, return the common prefix substring up to the current index.
6. If the loop reaches the end of the first string, then the whole first string is the common prefix. Return the entire first string. | LeetCode 14: Longest Common Prefix
longest-common-prefix
Write a function to find the longest common prefix string amongst an array of strings.
If there is no common prefix, return an empty string `" "`.
**Example 1:**
**Input:** strs = \[ "flower ", "flow ", "flight "\]
**Output:** "fl "
**Example 2:**
**Input:** strs = \[ "dog ", "racecar ", "car "\]
**Output:** " "
**Explanation:** There is no common prefix among the input strings.
**Constraints:**
* `1 <= strs.length <= 200`
* `0 <= strs[i].length <= 200`
* `strs[i]` consists of only lowercase English letters. | Solve the following problem in java, and explain your solution. |
```javascript
function longestCommonPrefix(strs) {
if (strs.length === 0) return "";
for (let i = 0; i < strs[0].length; ++i) {
const c = strs[0][i];
for (let j = 1; j < strs.length; ++j) {
if (i === strs[j].length || strs[j][i] !== c) {
return strs[0].substr(0, i);
}
}
}
return strs[0];
}
```
Explanation:
The algorithm iterates through the characters in the first string and compares the equivalent characters in the subsequent strings. If there is a mismatch detected, return the common prefix up to the mismatch point. If the algorithm finishes iterating over the first string, return the entire first string as the common prefix.
1. Check if the input array is empty, if it is, return an empty string as there is no common prefix.
2. Iterate through the characters of the first string.
3. For each character, iterate through the rest of the strings in the array.
4. Compare the current character with the corresponding character in the other strings.
5. If there is a mismatch or if we reach the end of a string, return the common prefix substring up to the current index.
6. If the loop reaches the end of the first string, then the whole first string is the common prefix. Return the entire first string. | LeetCode 14: Longest Common Prefix
longest-common-prefix
Write a function to find the longest common prefix string amongst an array of strings.
If there is no common prefix, return an empty string `" "`.
**Example 1:**
**Input:** strs = \[ "flower ", "flow ", "flight "\]
**Output:** "fl "
**Example 2:**
**Input:** strs = \[ "dog ", "racecar ", "car "\]
**Output:** " "
**Explanation:** There is no common prefix among the input strings.
**Constraints:**
* `1 <= strs.length <= 200`
* `0 <= strs[i].length <= 200`
* `strs[i]` consists of only lowercase English letters. | Solve the following problem in javascript, and explain your solution. |
```python
def longest_common_prefix(strs):
if not strs:
return ""
for i, c in enumerate(strs[0]):
for j in range(1, len(strs)):
if i == len(strs[j]) or strs[j][i] != c:
return strs[0][:i]
return strs[0]
```
Explanation:
The algorithm iterates through the characters in the first string and compares the equivalent characters in the subsequent strings. If there is a mismatch detected, return the common prefix up to the mismatch point. If the algorithm finishes iterating over the first string, return the entire first string as the common prefix.
1. Check if the input array is empty, if it is, return an empty string as there is no common prefix.
2. Iterate through the characters of the first string.
3. For each character, iterate through the rest of the strings in the array.
4. Compare the current character with the corresponding character in the other strings.
5. If there is a mismatch or if we reach the end of a string, return the common prefix substring up to the current index.
6. If the loop reaches the end of the first string, then the whole first string is the common prefix. Return the entire first string. | LeetCode 14: Longest Common Prefix
longest-common-prefix
Write a function to find the longest common prefix string amongst an array of strings.
If there is no common prefix, return an empty string `" "`.
**Example 1:**
**Input:** strs = \[ "flower ", "flow ", "flight "\]
**Output:** "fl "
**Example 2:**
**Input:** strs = \[ "dog ", "racecar ", "car "\]
**Output:** " "
**Explanation:** There is no common prefix among the input strings.
**Constraints:**
* `1 <= strs.length <= 200`
* `0 <= strs[i].length <= 200`
* `strs[i]` consists of only lowercase English letters. | Solve the following problem in python, and explain your solution. |
```cpp
#include <vector>
#include <algorithm>
std::vector<std::vector<int>> threeSum(std::vector<int>& nums) {
std::vector<std::vector<int>> result;
std::sort(nums.begin(), nums.end());
for (int i = 0; i < (int)nums.size() - 2; ++i) {
if (i == 0 || nums[i] != nums[i - 1]) {
int j = i + 1, k = nums.size() - 1;
while (j < k) {
int sum = nums[i] + nums[j] + nums[k];
if (sum == 0) {
result.push_back({nums[i], nums[j], nums[k]});
while (j < k && nums[j] == nums[j + 1]) ++j;
while (j < k && nums[k] == nums[k - 1]) --k;
++j;
--k;
} else if (sum < 0) {
++j;
} else {
--k;
}
}
}
}
return result;
}
```
Explanation:
1. Sort the input array `nums`.
2. Loop through `nums` from index `0` to `length - 2`. Let the current index be `i`.
- If `i` is `0` or `nums[i]` is not equal to the previous element (`nums[i - 1]`), perform the following steps:
- Initialize two pointers `j` and `k`, where `j` is the next index after `i`, and `k` is the last index of `nums`.
- While `j` is less than `k`, perform the following steps:
- Calculate the sum `s` of the elements at indices `i`, `j`, and `k`.
- If the sum `s` is equal to `0`, then we've found a valid triplet. Add it to the `result` array and move the pointers, `j` and `k`, skipping duplicates if any.
- If the sum `s` is less than `0`, increment `j` to move towards positive values.
- If the sum `s` is greater than `0`, decrement `k` to move towards negative values.
3. Return the `result` array.
The algorithm removes duplicates by checking if the current numbers being processed are equal to previously processed numbers. The sorting step helps in simplifying the algorithm and reducing the search space. | LeetCode 15: 3Sum
3sum
Given an integer array nums, return all the triplets `[nums[i], nums[j], nums[k]]` such that `i != j`, `i != k`, and `j != k`, and `nums[i] + nums[j] + nums[k] == 0`.
Notice that the solution set must not contain duplicate triplets.
**Example 1:**
**Input:** nums = \[-1,0,1,2,-1,-4\]
**Output:** \[\[-1,-1,2\],\[-1,0,1\]\]
**Explanation:**
nums\[0\] + nums\[1\] + nums\[2\] = (-1) + 0 + 1 = 0.
nums\[1\] + nums\[2\] + nums\[4\] = 0 + 1 + (-1) = 0.
nums\[0\] + nums\[3\] + nums\[4\] = (-1) + 2 + (-1) = 0.
The distinct triplets are \[-1,0,1\] and \[-1,-1,2\].
Notice that the order of the output and the order of the triplets does not matter.
**Example 2:**
**Input:** nums = \[0,1,1\]
**Output:** \[\]
**Explanation:** The only possible triplet does not sum up to 0.
**Example 3:**
**Input:** nums = \[0,0,0\]
**Output:** \[\[0,0,0\]\]
**Explanation:** The only possible triplet sums up to 0.
**Constraints:**
* `3 <= nums.length <= 3000`
* `-105 <= nums[i] <= 105` | Solve the following problem in c++, and explain your solution. |
```java
import java.util.Arrays;
import java.util.ArrayList;
import java.util.List;
public List<List<Integer>> threeSum(int[] nums) {
Arrays.sort(nums);
List<List<Integer>> result = new ArrayList<>();
for (int i = 0; i < nums.length - 2; i++) {
if (i == 0 || nums[i] != nums[i - 1]) {
int j = i + 1, k = nums.length - 1;
while (j < k) {
int sum = nums[i] + nums[j] + nums[k];
if (sum == 0) {
result.add(Arrays.asList(nums[i], nums[j], nums[k]));
while (j < k && nums[j] == nums[j + 1]) j++;
while (j < k && nums[k] == nums[k - 1]) k--;
j++;
k--;
} else if (sum < 0) {
j++;
} else {
k--;
}
}
}
}
return result;
}
```
Explanation:
1. Sort the input array `nums`.
2. Loop through `nums` from index `0` to `length - 2`. Let the current index be `i`.
- If `i` is `0` or `nums[i]` is not equal to the previous element (`nums[i - 1]`), perform the following steps:
- Initialize two pointers `j` and `k`, where `j` is the next index after `i`, and `k` is the last index of `nums`.
- While `j` is less than `k`, perform the following steps:
- Calculate the sum `s` of the elements at indices `i`, `j`, and `k`.
- If the sum `s` is equal to `0`, then we've found a valid triplet. Add it to the `result` array and move the pointers, `j` and `k`, skipping duplicates if any.
- If the sum `s` is less than `0`, increment `j` to move towards positive values.
- If the sum `s` is greater than `0`, decrement `k` to move towards negative values.
3. Return the `result` array.
The algorithm removes duplicates by checking if the current numbers being processed are equal to previously processed numbers. The sorting step helps in simplifying the algorithm and reducing the search space. | LeetCode 15: 3Sum
3sum
Given an integer array nums, return all the triplets `[nums[i], nums[j], nums[k]]` such that `i != j`, `i != k`, and `j != k`, and `nums[i] + nums[j] + nums[k] == 0`.
Notice that the solution set must not contain duplicate triplets.
**Example 1:**
**Input:** nums = \[-1,0,1,2,-1,-4\]
**Output:** \[\[-1,-1,2\],\[-1,0,1\]\]
**Explanation:**
nums\[0\] + nums\[1\] + nums\[2\] = (-1) + 0 + 1 = 0.
nums\[1\] + nums\[2\] + nums\[4\] = 0 + 1 + (-1) = 0.
nums\[0\] + nums\[3\] + nums\[4\] = (-1) + 2 + (-1) = 0.
The distinct triplets are \[-1,0,1\] and \[-1,-1,2\].
Notice that the order of the output and the order of the triplets does not matter.
**Example 2:**
**Input:** nums = \[0,1,1\]
**Output:** \[\]
**Explanation:** The only possible triplet does not sum up to 0.
**Example 3:**
**Input:** nums = \[0,0,0\]
**Output:** \[\[0,0,0\]\]
**Explanation:** The only possible triplet sums up to 0.
**Constraints:**
* `3 <= nums.length <= 3000`
* `-105 <= nums[i] <= 105` | Solve the following problem in java, and explain your solution. |
```javascript
function threeSum(nums) {
nums.sort((a, b) => a - b);
const result = [];
for (let i = 0; i < nums.length - 2; i++) {
if (i == 0 || nums[i] != nums[i - 1]) {
let j = i + 1, k = nums.length - 1;
while (j < k) {
const sum = nums[i] + nums[j] + nums[k];
if (sum === 0) {
result.push([nums[i], nums[j], nums[k]]);
while (j < k && nums[j] === nums[j + 1]) j++;
while (j < k && nums[k] === nums[k - 1]) k--;
j++;
k--;
} else if (sum < 0) {
j++;
} else {
k--;
}
}
}
}
return result;
}
```
Explanation:
1. Sort the input array `nums`.
2. Loop through `nums` from index `0` to `length - 2`. Let the current index be `i`.
- If `i` is `0` or `nums[i]` is not equal to the previous element (`nums[i - 1]`), perform the following steps:
- Initialize two pointers `j` and `k`, where `j` is the next index after `i`, and `k` is the last index of `nums`.
- While `j` is less than `k`, perform the following steps:
- Calculate the sum `s` of the elements at indices `i`, `j`, and `k`.
- If the sum `s` is equal to `0`, then we've found a valid triplet. Add it to the `result` array and move the pointers, `j` and `k`, skipping duplicates if any.
- If the sum `s` is less than `0`, increment `j` to move towards positive values.
- If the sum `s` is greater than `0`, decrement `k` to move towards negative values.
3. Return the `result` array.
The algorithm removes duplicates by checking if the current numbers being processed are equal to previously processed numbers. The sorting step helps in simplifying the algorithm and reducing the search space. | LeetCode 15: 3Sum
3sum
Given an integer array nums, return all the triplets `[nums[i], nums[j], nums[k]]` such that `i != j`, `i != k`, and `j != k`, and `nums[i] + nums[j] + nums[k] == 0`.
Notice that the solution set must not contain duplicate triplets.
**Example 1:**
**Input:** nums = \[-1,0,1,2,-1,-4\]
**Output:** \[\[-1,-1,2\],\[-1,0,1\]\]
**Explanation:**
nums\[0\] + nums\[1\] + nums\[2\] = (-1) + 0 + 1 = 0.
nums\[1\] + nums\[2\] + nums\[4\] = 0 + 1 + (-1) = 0.
nums\[0\] + nums\[3\] + nums\[4\] = (-1) + 2 + (-1) = 0.
The distinct triplets are \[-1,0,1\] and \[-1,-1,2\].
Notice that the order of the output and the order of the triplets does not matter.
**Example 2:**
**Input:** nums = \[0,1,1\]
**Output:** \[\]
**Explanation:** The only possible triplet does not sum up to 0.
**Example 3:**
**Input:** nums = \[0,0,0\]
**Output:** \[\[0,0,0\]\]
**Explanation:** The only possible triplet sums up to 0.
**Constraints:**
* `3 <= nums.length <= 3000`
* `-105 <= nums[i] <= 105` | Solve the following problem in javascript, and explain your solution. |
```python
def threeSum(nums):
nums.sort()
result = []
for i in range(len(nums) - 2):
if i == 0 or nums[i] != nums[i - 1]:
j, k = i + 1, len(nums) - 1
while j < k:
s = nums[i] + nums[j] + nums[k]
if s == 0:
result.append([nums[i], nums[j], nums[k]])
while j < k and nums[j] == nums[j + 1]:
j += 1
while j < k and nums[k] == nums[k - 1]:
k -= 1
j += 1
k -= 1
elif s < 0:
j += 1
else:
k -= 1
return result
```
Explanation:
1. Sort the input array `nums`.
2. Loop through `nums` from index `0` to `length - 2`. Let the current index be `i`.
- If `i` is `0` or `nums[i]` is not equal to the previous element (`nums[i - 1]`), perform the following steps:
- Initialize two pointers `j` and `k`, where `j` is the next index after `i`, and `k` is the last index of `nums`.
- While `j` is less than `k`, perform the following steps:
- Calculate the sum `s` of the elements at indices `i`, `j`, and `k`.
- If the sum `s` is equal to `0`, then we've found a valid triplet. Add it to the `result` array and move the pointers, `j` and `k`, skipping duplicates if any.
- If the sum `s` is less than `0`, increment `j` to move towards positive values.
- If the sum `s` is greater than `0`, decrement `k` to move towards negative values.
3. Return the `result` array.
The algorithm removes duplicates by checking if the current numbers being processed are equal to previously processed numbers. The sorting step helps in simplifying the algorithm and reducing the search space. | LeetCode 15: 3Sum
3sum
Given an integer array nums, return all the triplets `[nums[i], nums[j], nums[k]]` such that `i != j`, `i != k`, and `j != k`, and `nums[i] + nums[j] + nums[k] == 0`.
Notice that the solution set must not contain duplicate triplets.
**Example 1:**
**Input:** nums = \[-1,0,1,2,-1,-4\]
**Output:** \[\[-1,-1,2\],\[-1,0,1\]\]
**Explanation:**
nums\[0\] + nums\[1\] + nums\[2\] = (-1) + 0 + 1 = 0.
nums\[1\] + nums\[2\] + nums\[4\] = 0 + 1 + (-1) = 0.
nums\[0\] + nums\[3\] + nums\[4\] = (-1) + 2 + (-1) = 0.
The distinct triplets are \[-1,0,1\] and \[-1,-1,2\].
Notice that the order of the output and the order of the triplets does not matter.
**Example 2:**
**Input:** nums = \[0,1,1\]
**Output:** \[\]
**Explanation:** The only possible triplet does not sum up to 0.
**Example 3:**
**Input:** nums = \[0,0,0\]
**Output:** \[\[0,0,0\]\]
**Explanation:** The only possible triplet sums up to 0.
**Constraints:**
* `3 <= nums.length <= 3000`
* `-105 <= nums[i] <= 105` | Solve the following problem in python, and explain your solution. |
```cpp
#include <algorithm>
#include <vector>
int threeSumClosest(std::vector<int>& nums, int target) {
std::sort(nums.begin(), nums.end());
int closest = nums[0] + nums[1] + nums[2];
for (int i = 0; i < nums.size() - 2; ++i) {
int left = i + 1;
int right = nums.size() - 1;
while (left < right) {
int sum = nums[i] + nums[left] + nums[right];
if (sum == target) {
return sum;
}
if (abs(target - sum) < abs(target - closest)) {
closest = sum;
}
if (sum < target) {
left++;
} else {
right--;
}
}
}
return closest;
}
```
Explanation:
1. Sort the input array `nums`.
2. Initialize the `closest` variable to be the sum of the first three elements.
3. Iterate through the sorted array with a pointer `i` running from the first element to the third-to-last element.
4. Initialize two-pointers `left` (set to `i + 1`) and `right` (set to the last element).
5. While `left` is less than `right`:
a. Calculate the current sum `cur_sum` using the elements at positions `i`, `left`, and `right`.
b. If `cur_sum` is equal to `target`, return it as the closest sum.
c. Update the `closest` sum if the difference between `target` and `cur_sum` is less than the difference between `target` and `closest`.
d. Move the `left` pointer forward if `cur_sum` is less than `target`, otherwise move the `right` pointer backward.
6. Return the `closest` sum found. | LeetCode 16: 3Sum Closest
3sum-closest
Given an integer array `nums` of length `n` and an integer `target`, find three integers in `nums` such that the sum is closest to `target`.
Return _the sum of the three integers_.
You may assume that each input would have exactly one solution.
**Example 1:**
**Input:** nums = \[-1,2,1,-4\], target = 1
**Output:** 2
**Explanation:** The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
**Example 2:**
**Input:** nums = \[0,0,0\], target = 1
**Output:** 0
**Explanation:** The sum that is closest to the target is 0. (0 + 0 + 0 = 0).
**Constraints:**
* `3 <= nums.length <= 500`
* `-1000 <= nums[i] <= 1000`
* `-104 <= target <= 104` | Solve the following problem in c++, and explain your solution. |
```java
import java.util.Arrays;
public int threeSumClosest(int[] nums, int target) {
Arrays.sort(nums);
int closest = nums[0] + nums[1] + nums[2];
for (int i = 0; i < nums.length - 2; ++i) {
int left = i + 1;
int right = nums.length - 1;
while (left < right) {
int sum = nums[i] + nums[left] + nums[right];
if (sum == target) {
return sum;
}
if (Math.abs(target - sum) < Math.abs(target - closest)) {
closest = sum;
}
if (sum < target) {
left++;
} else {
right--;
}
}
}
return closest;
}
```
Explanation:
1. Sort the input array `nums`.
2. Initialize the `closest` variable to be the sum of the first three elements.
3. Iterate through the sorted array with a pointer `i` running from the first element to the third-to-last element.
4. Initialize two-pointers `left` (set to `i + 1`) and `right` (set to the last element).
5. While `left` is less than `right`:
a. Calculate the current sum `cur_sum` using the elements at positions `i`, `left`, and `right`.
b. If `cur_sum` is equal to `target`, return it as the closest sum.
c. Update the `closest` sum if the difference between `target` and `cur_sum` is less than the difference between `target` and `closest`.
d. Move the `left` pointer forward if `cur_sum` is less than `target`, otherwise move the `right` pointer backward.
6. Return the `closest` sum found. | LeetCode 16: 3Sum Closest
3sum-closest
Given an integer array `nums` of length `n` and an integer `target`, find three integers in `nums` such that the sum is closest to `target`.
Return _the sum of the three integers_.
You may assume that each input would have exactly one solution.
**Example 1:**
**Input:** nums = \[-1,2,1,-4\], target = 1
**Output:** 2
**Explanation:** The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
**Example 2:**
**Input:** nums = \[0,0,0\], target = 1
**Output:** 0
**Explanation:** The sum that is closest to the target is 0. (0 + 0 + 0 = 0).
**Constraints:**
* `3 <= nums.length <= 500`
* `-1000 <= nums[i] <= 1000`
* `-104 <= target <= 104` | Solve the following problem in java, and explain your solution. |
```javascript
function threeSumClosest(nums, target) {
nums.sort((a, b) => a - b);
let closest = nums[0] + nums[1] + nums[2];
for (let i = 0; i < nums.length - 2; ++i) {
let left = i + 1;
let right = nums.length - 1;
while (left < right) {
let sum = nums[i] + nums[left] + nums[right];
if (sum === target) {
return sum;
}
if (Math.abs(target - sum) < Math.abs(target - closest)) {
closest = sum;
}
if (sum < target) {
left++;
} else {
right--;
}
}
}
return closest;
}
```
Explanation:
1. Sort the input array `nums`.
2. Initialize the `closest` variable to be the sum of the first three elements.
3. Iterate through the sorted array with a pointer `i` running from the first element to the third-to-last element.
4. Initialize two-pointers `left` (set to `i + 1`) and `right` (set to the last element).
5. While `left` is less than `right`:
a. Calculate the current sum `cur_sum` using the elements at positions `i`, `left`, and `right`.
b. If `cur_sum` is equal to `target`, return it as the closest sum.
c. Update the `closest` sum if the difference between `target` and `cur_sum` is less than the difference between `target` and `closest`.
d. Move the `left` pointer forward if `cur_sum` is less than `target`, otherwise move the `right` pointer backward.
6. Return the `closest` sum found. | LeetCode 16: 3Sum Closest
3sum-closest
Given an integer array `nums` of length `n` and an integer `target`, find three integers in `nums` such that the sum is closest to `target`.
Return _the sum of the three integers_.
You may assume that each input would have exactly one solution.
**Example 1:**
**Input:** nums = \[-1,2,1,-4\], target = 1
**Output:** 2
**Explanation:** The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
**Example 2:**
**Input:** nums = \[0,0,0\], target = 1
**Output:** 0
**Explanation:** The sum that is closest to the target is 0. (0 + 0 + 0 = 0).
**Constraints:**
* `3 <= nums.length <= 500`
* `-1000 <= nums[i] <= 1000`
* `-104 <= target <= 104` | Solve the following problem in javascript, and explain your solution. |
```python
def threeSumClosest(nums, target):
nums.sort()
closest = sum(nums[:3])
for i in range(len(nums) - 2):
left, right = i + 1, len(nums) - 1
while left < right:
cur_sum = nums[i] + nums[left] + nums[right]
if cur_sum == target:
return cur_sum
if abs(target - cur_sum) < abs(target - closest):
closest = cur_sum
if cur_sum < target:
left += 1
else:
right -= 1
return closest
```
Explanation:
1. Sort the input array `nums`.
2. Initialize the `closest` variable to be the sum of the first three elements.
3. Iterate through the sorted array with a pointer `i` running from the first element to the third-to-last element.
4. Initialize two-pointers `left` (set to `i + 1`) and `right` (set to the last element).
5. While `left` is less than `right`:
a. Calculate the current sum `cur_sum` using the elements at positions `i`, `left`, and `right`.
b. If `cur_sum` is equal to `target`, return it as the closest sum.
c. Update the `closest` sum if the difference between `target` and `cur_sum` is less than the difference between `target` and `closest`.
d. Move the `left` pointer forward if `cur_sum` is less than `target`, otherwise move the `right` pointer backward.
6. Return the `closest` sum found. | LeetCode 16: 3Sum Closest
3sum-closest
Given an integer array `nums` of length `n` and an integer `target`, find three integers in `nums` such that the sum is closest to `target`.
Return _the sum of the three integers_.
You may assume that each input would have exactly one solution.
**Example 1:**
**Input:** nums = \[-1,2,1,-4\], target = 1
**Output:** 2
**Explanation:** The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
**Example 2:**
**Input:** nums = \[0,0,0\], target = 1
**Output:** 0
**Explanation:** The sum that is closest to the target is 0. (0 + 0 + 0 = 0).
**Constraints:**
* `3 <= nums.length <= 500`
* `-1000 <= nums[i] <= 1000`
* `-104 <= target <= 104` | Solve the following problem in python, and explain your solution. |
```cpp
#include <vector>
#include <string>
using namespace std;
vector<string> letterCombinations(string digits) {
if(digits.empty()) return {};
vector<string> phone = {"abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
vector<string> result = {""};
for (char d : digits) {
vector<string> temp;
for (const string &s : result) {
for (char c : phone[d - '2']) {
temp.push_back(s + c);
}
}
result.swap(temp);
}
return result;
}
```
Explanation:
The algorithm visits each digit from the given `digits` string one by one. For each digit, it appends all possible characters corresponding to that digit, in a breadth-first search manner. The current stage's results are saved in the `result` list (or `output` list in Java). A temporary list `temp` is used to store the combinations formed at each stage before replacing the `result` list. The phone buttons mapping is stored in the `phone` list/array. | LeetCode 17: Letter Combinations of a Phone Number
letter-combinations-of-a-phone-number
Given a string containing digits from `2-9` inclusive, return all possible letter combinations that the number could represent. Return the answer in **any order**.
A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.
**Example 1:**
**Input:** digits = "23 "
**Output:** \[ "ad ", "ae ", "af ", "bd ", "be ", "bf ", "cd ", "ce ", "cf "\]
**Example 2:**
**Input:** digits = " "
**Output:** \[\]
**Example 3:**
**Input:** digits = "2 "
**Output:** \[ "a ", "b ", "c "\]
**Constraints:**
* `0 <= digits.length <= 4`
* `digits[i]` is a digit in the range `['2', '9']`. | Solve the following problem in c++, and explain your solution. |
```java
public List<String> letterCombinations(String digits) {
LinkedList<String> output = new LinkedList<>();
if(digits.isEmpty()) return output;
String[] phone = new String[] {"abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
output.add("");
for(char d : digits.toCharArray()){
while(output.peek().length() == digits.indexOf(d)){
String perm = output.remove();
for(char c : phone[d - '2'].toCharArray()){
output.add(perm + c);
}
}
}
return output;
}
```
Explanation:
The algorithm visits each digit from the given `digits` string one by one. For each digit, it appends all possible characters corresponding to that digit, in a breadth-first search manner. The current stage's results are saved in the `result` list (or `output` list in Java). A temporary list `temp` is used to store the combinations formed at each stage before replacing the `result` list. The phone buttons mapping is stored in the `phone` list/array. | LeetCode 17: Letter Combinations of a Phone Number
letter-combinations-of-a-phone-number
Given a string containing digits from `2-9` inclusive, return all possible letter combinations that the number could represent. Return the answer in **any order**.
A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.
**Example 1:**
**Input:** digits = "23 "
**Output:** \[ "ad ", "ae ", "af ", "bd ", "be ", "bf ", "cd ", "ce ", "cf "\]
**Example 2:**
**Input:** digits = " "
**Output:** \[\]
**Example 3:**
**Input:** digits = "2 "
**Output:** \[ "a ", "b ", "c "\]
**Constraints:**
* `0 <= digits.length <= 4`
* `digits[i]` is a digit in the range `['2', '9']`. | Solve the following problem in java, and explain your solution. |
```javascript
function letterCombinations(digits) {
if (digits.length === 0) return [];
let phone = ["abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"];
let result = [""];
for (let i = 0; i < digits.length; i++) {
let digit = digits[i];
let temp = [];
for (let s of result) {
for (let c of phone[digit - 2]) {
temp.push(s + c);
}
}
result = temp;
}
return result;
}
```
Explanation:
The algorithm visits each digit from the given `digits` string one by one. For each digit, it appends all possible characters corresponding to that digit, in a breadth-first search manner. The current stage's results are saved in the `result` list (or `output` list in Java). A temporary list `temp` is used to store the combinations formed at each stage before replacing the `result` list. The phone buttons mapping is stored in the `phone` list/array. | LeetCode 17: Letter Combinations of a Phone Number
letter-combinations-of-a-phone-number
Given a string containing digits from `2-9` inclusive, return all possible letter combinations that the number could represent. Return the answer in **any order**.
A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.
**Example 1:**
**Input:** digits = "23 "
**Output:** \[ "ad ", "ae ", "af ", "bd ", "be ", "bf ", "cd ", "ce ", "cf "\]
**Example 2:**
**Input:** digits = " "
**Output:** \[\]
**Example 3:**
**Input:** digits = "2 "
**Output:** \[ "a ", "b ", "c "\]
**Constraints:**
* `0 <= digits.length <= 4`
* `digits[i]` is a digit in the range `['2', '9']`. | Solve the following problem in javascript, and explain your solution. |
```python
def letter_combinations(digits: str):
if not digits: return []
phone = ["abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"]
result = [""]
for digit in digits:
temp = []
for s in result:
for c in phone[int(digit) - 2]:
temp.append(s + c)
result = temp
return result
```
Explanation:
The algorithm visits each digit from the given `digits` string one by one. For each digit, it appends all possible characters corresponding to that digit, in a breadth-first search manner. The current stage's results are saved in the `result` list (or `output` list in Java). A temporary list `temp` is used to store the combinations formed at each stage before replacing the `result` list. The phone buttons mapping is stored in the `phone` list/array. | LeetCode 17: Letter Combinations of a Phone Number
letter-combinations-of-a-phone-number
Given a string containing digits from `2-9` inclusive, return all possible letter combinations that the number could represent. Return the answer in **any order**.
A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.
**Example 1:**
**Input:** digits = "23 "
**Output:** \[ "ad ", "ae ", "af ", "bd ", "be ", "bf ", "cd ", "ce ", "cf "\]
**Example 2:**
**Input:** digits = " "
**Output:** \[\]
**Example 3:**
**Input:** digits = "2 "
**Output:** \[ "a ", "b ", "c "\]
**Constraints:**
* `0 <= digits.length <= 4`
* `digits[i]` is a digit in the range `['2', '9']`. | Solve the following problem in python, and explain your solution. |
```cpp
#include <vector>
#include <algorithm>
using namespace std;
vector<vector<int>> fourSum(vector<int>& nums, int target) {
vector<vector<int>> result;
int n = nums.size();
if(n < 4)
return result;
sort(nums.begin(), nums.end());
for(int i = 0; i < n - 3; i++) {
if(i > 0 && nums[i] == nums[i-1])
continue;
for(int j = i + 1; j < n - 2; j++) {
if(j > i + 1 && nums[j] == nums[j-1])
continue;
int left = j + 1;
int right = n - 1;
while(left < right) {
int sum = nums[i] + nums[j] + nums[left] + nums[right];
if(sum == target) {
result.push_back({nums[i], nums[j], nums[left], nums[right]});
++left;
--right;
while(left < right && nums[left] == nums[left-1]) ++left;
while(left < right && nums[right] == nums[right+1]) --right;
} else if(sum < target) {
++left;
} else {
--right;
}
}
}
}
return result;
}
```
Explanation:
1. First, sort the given array of n integers.
2. Iterate through the array with two nested loops to choose two (not necessarily distinct) elements (i, j) where i < j.
3. Create two pointers, left and right, starting from j + 1 and n - 1, respectively, to find a pair of elements (left, right) with sum equal to target - (nums[i] + nums[j]).
4. Check whether the current quadruplet [nums[i], nums[j], nums[left], nums[right]] is unique. If the same quadruplet is not already in the result, add it to the result.
5. To avoid repetition, if the next element is the same as the current element in either i, j, left, or right, skip the current iteration and move to the next one.
6. Return the resulting array of unique quadruplets. | LeetCode 18: 4Sum
4sum
Given an array `nums` of `n` integers, return _an array of all the **unique** quadruplets_ `[nums[a], nums[b], nums[c], nums[d]]` such that:
* `0 <= a, b, c, d < n`
* `a`, `b`, `c`, and `d` are **distinct**.
* `nums[a] + nums[b] + nums[c] + nums[d] == target`
You may return the answer in **any order**.
**Example 1:**
**Input:** nums = \[1,0,-1,0,-2,2\], target = 0
**Output:** \[\[-2,-1,1,2\],\[-2,0,0,2\],\[-1,0,0,1\]\]
**Example 2:**
**Input:** nums = \[2,2,2,2,2\], target = 8
**Output:** \[\[2,2,2,2\]\]
**Constraints:**
* `1 <= nums.length <= 200`
* `-109 <= nums[i] <= 109`
* `-109 <= target <= 109` | Solve the following problem in c++, and explain your solution. |
```java
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
public List<List<Integer>> fourSum(int[] nums, int target) {
List<List<Integer>> result = new ArrayList<>();
if(nums == null || nums.length < 4)
return result;
Arrays.sort(nums);
for(int i = 0; i < nums.length - 3; i++) {
if(i > 0 && nums[i] == nums[i-1])
continue;
for(int j = i + 1; j < nums.length - 2; j++) {
if(j > i + 1 && nums[j] == nums[j-1])
continue;
int left = j + 1;
int right = nums.length - 1;
while(left < right) {
int sum = nums[i] + nums[j] + nums[left] + nums[right];
if(sum == target) {
result.add(Arrays.asList(nums[i], nums[j], nums[left], nums[right]));
++left;
--right;
while(left < right && nums[left] == nums[left - 1]) ++left;
while(left < right && nums[right] == nums[right + 1]) --right;
} else if(sum < target) {
++left;
} else {
--right;
}
}
}
}
return result;
}
```
Explanation:
1. First, sort the given array of n integers.
2. Iterate through the array with two nested loops to choose two (not necessarily distinct) elements (i, j) where i < j.
3. Create two pointers, left and right, starting from j + 1 and n - 1, respectively, to find a pair of elements (left, right) with sum equal to target - (nums[i] + nums[j]).
4. Check whether the current quadruplet [nums[i], nums[j], nums[left], nums[right]] is unique. If the same quadruplet is not already in the result, add it to the result.
5. To avoid repetition, if the next element is the same as the current element in either i, j, left, or right, skip the current iteration and move to the next one.
6. Return the resulting array of unique quadruplets. | LeetCode 18: 4Sum
4sum
Given an array `nums` of `n` integers, return _an array of all the **unique** quadruplets_ `[nums[a], nums[b], nums[c], nums[d]]` such that:
* `0 <= a, b, c, d < n`
* `a`, `b`, `c`, and `d` are **distinct**.
* `nums[a] + nums[b] + nums[c] + nums[d] == target`
You may return the answer in **any order**.
**Example 1:**
**Input:** nums = \[1,0,-1,0,-2,2\], target = 0
**Output:** \[\[-2,-1,1,2\],\[-2,0,0,2\],\[-1,0,0,1\]\]
**Example 2:**
**Input:** nums = \[2,2,2,2,2\], target = 8
**Output:** \[\[2,2,2,2\]\]
**Constraints:**
* `1 <= nums.length <= 200`
* `-109 <= nums[i] <= 109`
* `-109 <= target <= 109` | Solve the following problem in java, and explain your solution. |
```javascript
function fourSum(nums, target) {
nums.sort((a, b) => a - b);
let result = [];
for (let i = 0; i < nums.length - 3; i++) {
if (i > 0 && nums[i] === nums[i - 1]) continue;
for (let j = i + 1; j < nums.length - 2; j++) {
if (j > i + 1 && nums[j] === nums[j - 1]) continue;
let left = j + 1;
let right = nums.length - 1;
while (left < right) {
const sum = nums[i] + nums[j] + nums[left] + nums[right];
if (sum === target) {
result.push([nums[i], nums[j], nums[left], nums[right]]);
left++;
right--;
while (left < right && nums[left] === nums[left - 1]) left++;
while (left < right && nums[right] === nums[right + 1]) right--;
} else if (sum < target) {
left++;
} else {
right--;
}
}
}
}
return result;
}
```
Explanation:
1. First, sort the given array of n integers.
2. Iterate through the array with two nested loops to choose two (not necessarily distinct) elements (i, j) where i < j.
3. Create two pointers, left and right, starting from j + 1 and n - 1, respectively, to find a pair of elements (left, right) with sum equal to target - (nums[i] + nums[j]).
4. Check whether the current quadruplet [nums[i], nums[j], nums[left], nums[right]] is unique. If the same quadruplet is not already in the result, add it to the result.
5. To avoid repetition, if the next element is the same as the current element in either i, j, left, or right, skip the current iteration and move to the next one.
6. Return the resulting array of unique quadruplets. | LeetCode 18: 4Sum
4sum
Given an array `nums` of `n` integers, return _an array of all the **unique** quadruplets_ `[nums[a], nums[b], nums[c], nums[d]]` such that:
* `0 <= a, b, c, d < n`
* `a`, `b`, `c`, and `d` are **distinct**.
* `nums[a] + nums[b] + nums[c] + nums[d] == target`
You may return the answer in **any order**.
**Example 1:**
**Input:** nums = \[1,0,-1,0,-2,2\], target = 0
**Output:** \[\[-2,-1,1,2\],\[-2,0,0,2\],\[-1,0,0,1\]\]
**Example 2:**
**Input:** nums = \[2,2,2,2,2\], target = 8
**Output:** \[\[2,2,2,2\]\]
**Constraints:**
* `1 <= nums.length <= 200`
* `-109 <= nums[i] <= 109`
* `-109 <= target <= 109` | Solve the following problem in javascript, and explain your solution. |
```python
def fourSum(nums, target):
nums.sort()
result = []
for i in range(len(nums) - 3):
if i > 0 and nums[i] == nums[i - 1]:
continue
for j in range(i + 1, len(nums) - 2):
if j > i + 1 and nums[j] == nums[j - 1]:
continue
left = j + 1
right = len(nums) - 1
while left < right:
cur_sum = nums[i] + nums[j] + nums[left] + nums[right]
if cur_sum == target:
result.append([nums[i], nums[j], nums[left], nums[right]])
left += 1
right -= 1
while left < right and nums[left] == nums[left - 1]: left += 1
while left < right and nums[right] == nums[right + 1]: right -= 1
elif cur_sum < target:
left += 1
else:
right -= 1
return result
```
Explanation:
1. First, sort the given array of n integers.
2. Iterate through the array with two nested loops to choose two (not necessarily distinct) elements (i, j) where i < j.
3. Create two pointers, left and right, starting from j + 1 and n - 1, respectively, to find a pair of elements (left, right) with sum equal to target - (nums[i] + nums[j]).
4. Check whether the current quadruplet [nums[i], nums[j], nums[left], nums[right]] is unique. If the same quadruplet is not already in the result, add it to the result.
5. To avoid repetition, if the next element is the same as the current element in either i, j, left, or right, skip the current iteration and move to the next one.
6. Return the resulting array of unique quadruplets. | LeetCode 18: 4Sum
4sum
Given an array `nums` of `n` integers, return _an array of all the **unique** quadruplets_ `[nums[a], nums[b], nums[c], nums[d]]` such that:
* `0 <= a, b, c, d < n`
* `a`, `b`, `c`, and `d` are **distinct**.
* `nums[a] + nums[b] + nums[c] + nums[d] == target`
You may return the answer in **any order**.
**Example 1:**
**Input:** nums = \[1,0,-1,0,-2,2\], target = 0
**Output:** \[\[-2,-1,1,2\],\[-2,0,0,2\],\[-1,0,0,1\]\]
**Example 2:**
**Input:** nums = \[2,2,2,2,2\], target = 8
**Output:** \[\[2,2,2,2\]\]
**Constraints:**
* `1 <= nums.length <= 200`
* `-109 <= nums[i] <= 109`
* `-109 <= target <= 109` | Solve the following problem in python, and explain your solution. |
```cpp
class ListNode {
public:
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};
ListNode* removeNthFromEnd(ListNode* head, int n) {
ListNode* first = head;
ListNode* second = head;
for (int i = 0; i < n; i++) {
first = first->next;
}
if (first == NULL) {
head = head->next;
return head;
}
while (first->next != NULL) {
first = first->next;
second = second->next;
}
second->next = second->next->next;
return head;
}
```
Explanation:
The algorithm uses two pointers, first and second. The first pointer traverses the list, moving n steps ahead of the second pointer. If the first pointer reaches the end of the list, then the head is updated, and the second pointer becomes the new head. Otherwise, the second pointer is moved along with the first pointer until the first pointer is one position before the last item. Then, the nth node from the end of the list is skipped by updating the next pointer of the second pointer. Finally, the head of the updated list is returned. | LeetCode 19: Remove Nth Node From End of List
remove-nth-node-from-end-of-list
Given the `head` of a linked list, remove the `nth` node from the end of the list and return its head.
**Example 1:**
**Input:** head = \[1,2,3,4,5\], n = 2
**Output:** \[1,2,3,5\]
**Example 2:**
**Input:** head = \[1\], n = 1
**Output:** \[\]
**Example 3:**
**Input:** head = \[1,2\], n = 1
**Output:** \[1\]
**Constraints:**
* The number of nodes in the list is `sz`.
* `1 <= sz <= 30`
* `0 <= Node.val <= 100`
* `1 <= n <= sz`
**Follow up:** Could you do this in one pass? | Solve the following problem in c++, and explain your solution. |
```java
public class ListNode {
int val;
ListNode next;
ListNode(int x) { val = x; }
}
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode first = head;
ListNode second = head;
for (int i = 0; i < n; i++) {
first = first.next;
}
if (first == null) {
head = head.next;
return head;
}
while (first.next != null) {
first = first.next;
second = second.next;
}
second.next = second.next.next;
return head;
}
```
Explanation:
The algorithm uses two pointers, first and second. The first pointer traverses the list, moving n steps ahead of the second pointer. If the first pointer reaches the end of the list, then the head is updated, and the second pointer becomes the new head. Otherwise, the second pointer is moved along with the first pointer until the first pointer is one position before the last item. Then, the nth node from the end of the list is skipped by updating the next pointer of the second pointer. Finally, the head of the updated list is returned. | LeetCode 19: Remove Nth Node From End of List
remove-nth-node-from-end-of-list
Given the `head` of a linked list, remove the `nth` node from the end of the list and return its head.
**Example 1:**
**Input:** head = \[1,2,3,4,5\], n = 2
**Output:** \[1,2,3,5\]
**Example 2:**
**Input:** head = \[1\], n = 1
**Output:** \[\]
**Example 3:**
**Input:** head = \[1,2\], n = 1
**Output:** \[1\]
**Constraints:**
* The number of nodes in the list is `sz`.
* `1 <= sz <= 30`
* `0 <= Node.val <= 100`
* `1 <= n <= sz`
**Follow up:** Could you do this in one pass? | Solve the following problem in java, and explain your solution. |
```javascript
class ListNode {
constructor(val = 0, next = null) {
this.val = val;
this.next = next;
}
}
function removeNthFromEnd(head, n) {
let first = head;
let second = head;
for (let i = 0; i < n; i++) {
first = first.next;
}
if (first === null) {
head = head.next;
return head;
}
while (first.next !== null) {
first = first.next;
second = second.next;
}
second.next = second.next.next;
return head;
}
```
Explanation:
The algorithm uses two pointers, first and second. The first pointer traverses the list, moving n steps ahead of the second pointer. If the first pointer reaches the end of the list, then the head is updated, and the second pointer becomes the new head. Otherwise, the second pointer is moved along with the first pointer until the first pointer is one position before the last item. Then, the nth node from the end of the list is skipped by updating the next pointer of the second pointer. Finally, the head of the updated list is returned. | LeetCode 19: Remove Nth Node From End of List
remove-nth-node-from-end-of-list
Given the `head` of a linked list, remove the `nth` node from the end of the list and return its head.
**Example 1:**
**Input:** head = \[1,2,3,4,5\], n = 2
**Output:** \[1,2,3,5\]
**Example 2:**
**Input:** head = \[1\], n = 1
**Output:** \[\]
**Example 3:**
**Input:** head = \[1,2\], n = 1
**Output:** \[1\]
**Constraints:**
* The number of nodes in the list is `sz`.
* `1 <= sz <= 30`
* `0 <= Node.val <= 100`
* `1 <= n <= sz`
**Follow up:** Could you do this in one pass? | Solve the following problem in javascript, and explain your solution. |
```python
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
def removeNthFromEnd(head, n):
first = head
second = head
for i in range(n):
first = first.next
if not first:
head = head.next
return head
while first.next:
first = first.next
second = second.next
second.next = second.next.next
return head
```
Explanation:
The algorithm uses two pointers, first and second. The first pointer traverses the list, moving n steps ahead of the second pointer. If the first pointer reaches the end of the list, then the head is updated, and the second pointer becomes the new head. Otherwise, the second pointer is moved along with the first pointer until the first pointer is one position before the last item. Then, the nth node from the end of the list is skipped by updating the next pointer of the second pointer. Finally, the head of the updated list is returned. | LeetCode 19: Remove Nth Node From End of List
remove-nth-node-from-end-of-list
Given the `head` of a linked list, remove the `nth` node from the end of the list and return its head.
**Example 1:**
**Input:** head = \[1,2,3,4,5\], n = 2
**Output:** \[1,2,3,5\]
**Example 2:**
**Input:** head = \[1\], n = 1
**Output:** \[\]
**Example 3:**
**Input:** head = \[1,2\], n = 1
**Output:** \[1\]
**Constraints:**
* The number of nodes in the list is `sz`.
* `1 <= sz <= 30`
* `0 <= Node.val <= 100`
* `1 <= n <= sz`
**Follow up:** Could you do this in one pass? | Solve the following problem in python, and explain your solution. |
```cpp
bool isValid(std::string s) {
std::stack<char> stack;
for (char c : s) {
if (c == '(' || c == '{' || c == '[') {
stack.push(c);
} else {
if (stack.empty()) return false;
if (c == ')' && stack.top() != '(') return false;
if (c == '}' && stack.top() != '{') return false;
if (c == ']' && stack.top() != '[') return false;
stack.pop();
}
}
return stack.empty();
}
```
Explanation:
The algorithm uses a stack data structure to check the validity of the input string. It iterates through the string one character at a time. When an open bracket is encountered, it is pushed onto the stack. When a close bracket is encountered, the algorithm checks if the stack is empty or the corresponding open bracket is not at the top of the stack. If either of these conditions is true, the function returns false. If not, the open bracket is popped from the stack. After iterating through the entire string, the function checks if the stack is empty. If the stack is not empty, the function returns false, indicating there were unmatched open braces. Otherwise, the function returns true, indicating that the string is valid. | LeetCode 20: Valid Parentheses
valid-parentheses
Given a string `s` containing just the characters `'('`, `')'`, `'{'`, `'}'`, `'['` and `']'`, determine if the input string is valid.
An input string is valid if:
1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.
3. Every close bracket has a corresponding open bracket of the same type.
**Example 1:**
**Input:** s = "() "
**Output:** true
**Example 2:**
**Input:** s = "()\[\]{} "
**Output:** true
**Example 3:**
**Input:** s = "(\] "
**Output:** false
**Constraints:**
* `1 <= s.length <= 104`
* `s` consists of parentheses only `'()[]{}'`. | Solve the following problem in c++, and explain your solution. |
```java
boolean isValid(String s) {
Stack<Character> stack = new Stack<>();
for (char c : s.toCharArray()) {
if (c == '(' || c == '{' || c == '[') {
stack.push(c);
} else {
if (stack.empty()) return false;
if (c == ')' && stack.peek() != '(') return false;
if (c == '}' && stack.peek() != '{') return false;
if (c == ']' && stack.peek() != '[') return false;
stack.pop();
}
}
return stack.empty();
}
```
Explanation:
The algorithm uses a stack data structure to check the validity of the input string. It iterates through the string one character at a time. When an open bracket is encountered, it is pushed onto the stack. When a close bracket is encountered, the algorithm checks if the stack is empty or the corresponding open bracket is not at the top of the stack. If either of these conditions is true, the function returns false. If not, the open bracket is popped from the stack. After iterating through the entire string, the function checks if the stack is empty. If the stack is not empty, the function returns false, indicating there were unmatched open braces. Otherwise, the function returns true, indicating that the string is valid. | LeetCode 20: Valid Parentheses
valid-parentheses
Given a string `s` containing just the characters `'('`, `')'`, `'{'`, `'}'`, `'['` and `']'`, determine if the input string is valid.
An input string is valid if:
1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.
3. Every close bracket has a corresponding open bracket of the same type.
**Example 1:**
**Input:** s = "() "
**Output:** true
**Example 2:**
**Input:** s = "()\[\]{} "
**Output:** true
**Example 3:**
**Input:** s = "(\] "
**Output:** false
**Constraints:**
* `1 <= s.length <= 104`
* `s` consists of parentheses only `'()[]{}'`. | Solve the following problem in java, and explain your solution. |
```javascript
function isValid(s) {
const stack = [];
for (const c of s) {
if (c === '(' || c === '{' || c === '[') {
stack.push(c);
} else {
if (!stack.length) return false;
if (c === ')' && stack[stack.length - 1] !== '(') return false;
if (c === '}' && stack[stack.length - 1] !== '{') return false;
if (c === ']' && stack[stack.length - 1] !== '[') return false;
stack.pop();
}
}
return stack.length === 0;
}
```
Explanation:
The algorithm uses a stack data structure to check the validity of the input string. It iterates through the string one character at a time. When an open bracket is encountered, it is pushed onto the stack. When a close bracket is encountered, the algorithm checks if the stack is empty or the corresponding open bracket is not at the top of the stack. If either of these conditions is true, the function returns false. If not, the open bracket is popped from the stack. After iterating through the entire string, the function checks if the stack is empty. If the stack is not empty, the function returns false, indicating there were unmatched open braces. Otherwise, the function returns true, indicating that the string is valid. | LeetCode 20: Valid Parentheses
valid-parentheses
Given a string `s` containing just the characters `'('`, `')'`, `'{'`, `'}'`, `'['` and `']'`, determine if the input string is valid.
An input string is valid if:
1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.
3. Every close bracket has a corresponding open bracket of the same type.
**Example 1:**
**Input:** s = "() "
**Output:** true
**Example 2:**
**Input:** s = "()\[\]{} "
**Output:** true
**Example 3:**
**Input:** s = "(\] "
**Output:** false
**Constraints:**
* `1 <= s.length <= 104`
* `s` consists of parentheses only `'()[]{}'`. | Solve the following problem in javascript, and explain your solution. |
```python
def is_valid(s: str) -> bool:
stack = []
for c in s:
if c in '([{':
stack.append(c)
else:
if not stack:
return False
if c == ')' and stack[-1] != '(':
return False
if c == '}' and stack[-1] != '{':
return False
if c == ']' and stack[-1] != '[':
return False
stack.pop()
return not stack
```
Explanation:
The algorithm uses a stack data structure to check the validity of the input string. It iterates through the string one character at a time. When an open bracket is encountered, it is pushed onto the stack. When a close bracket is encountered, the algorithm checks if the stack is empty or the corresponding open bracket is not at the top of the stack. If either of these conditions is true, the function returns false. If not, the open bracket is popped from the stack. After iterating through the entire string, the function checks if the stack is empty. If the stack is not empty, the function returns false, indicating there were unmatched open braces. Otherwise, the function returns true, indicating that the string is valid. | LeetCode 20: Valid Parentheses
valid-parentheses
Given a string `s` containing just the characters `'('`, `')'`, `'{'`, `'}'`, `'['` and `']'`, determine if the input string is valid.
An input string is valid if:
1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.
3. Every close bracket has a corresponding open bracket of the same type.
**Example 1:**
**Input:** s = "() "
**Output:** true
**Example 2:**
**Input:** s = "()\[\]{} "
**Output:** true
**Example 3:**
**Input:** s = "(\] "
**Output:** false
**Constraints:**
* `1 <= s.length <= 104`
* `s` consists of parentheses only `'()[]{}'`. | Solve the following problem in python, and explain your solution. |
```cpp
ListNode* mergeTwoLists(ListNode* list1, ListNode* list2) {
if (list1 == nullptr) return list2;
if (list2 == nullptr) return list1;
if (list1->val < list2->val) {
list1->next = mergeTwoLists(list1->next, list2);
return list1;
} else {
list2->next = mergeTwoLists(list1, list2->next);
return list2;
}
}
```
Explanation:
The algorithm is a recursive solution:
1. If either `list1` or `list2` is null, return the other list as the result.
2. Compare the values of the nodes in `list1` and `list2`.
3. If `list1` node value is smaller, set `list1.next` to the result of recursively calling the function with `list1.next` and `list2`.
4. Otherwise, set `list2.next` to the result of recursively calling the function with `list1` and `list2.next`.
5. Return the current node as the new head of the merged list.
This process is repeated until one of the lists becomes null, then the other list is returned as the remainder of the merged list.
Please note that each language has some differences in syntax, like defining a class in Python or using the `nullptr` keyword in C++. However, the core logic of the solution remains the same for all languages. | LeetCode 21: Merge Two Sorted Lists
merge-two-sorted-lists
You are given the heads of two sorted linked lists `list1` and `list2`.
Merge the two lists in a one **sorted** list. The list should be made by splicing together the nodes of the first two lists.
Return _the head of the merged linked list_.
**Example 1:**
**Input:** list1 = \[1,2,4\], list2 = \[1,3,4\]
**Output:** \[1,1,2,3,4,4\]
**Example 2:**
**Input:** list1 = \[\], list2 = \[\]
**Output:** \[\]
**Example 3:**
**Input:** list1 = \[\], list2 = \[0\]
**Output:** \[0\]
**Constraints:**
* The number of nodes in both lists is in the range `[0, 50]`.
* `-100 <= Node.val <= 100`
* Both `list1` and `list2` are sorted in **non-decreasing** order. | Solve the following problem in c++, and explain your solution. |
```java
public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
if (list1 == null) return list2;
if (list2 == null) return list1;
if (list1.val < list2.val) {
list1.next = mergeTwoLists(list1.next, list2);
return list1;
} else {
list2.next = mergeTwoLists(list1, list2.next);
return list2;
}
}
```
Explanation:
The algorithm is a recursive solution:
1. If either `list1` or `list2` is null, return the other list as the result.
2. Compare the values of the nodes in `list1` and `list2`.
3. If `list1` node value is smaller, set `list1.next` to the result of recursively calling the function with `list1.next` and `list2`.
4. Otherwise, set `list2.next` to the result of recursively calling the function with `list1` and `list2.next`.
5. Return the current node as the new head of the merged list.
This process is repeated until one of the lists becomes null, then the other list is returned as the remainder of the merged list.
Please note that each language has some differences in syntax, like defining a class in Python or using the `nullptr` keyword in C++. However, the core logic of the solution remains the same for all languages. | LeetCode 21: Merge Two Sorted Lists
merge-two-sorted-lists
You are given the heads of two sorted linked lists `list1` and `list2`.
Merge the two lists in a one **sorted** list. The list should be made by splicing together the nodes of the first two lists.
Return _the head of the merged linked list_.
**Example 1:**
**Input:** list1 = \[1,2,4\], list2 = \[1,3,4\]
**Output:** \[1,1,2,3,4,4\]
**Example 2:**
**Input:** list1 = \[\], list2 = \[\]
**Output:** \[\]
**Example 3:**
**Input:** list1 = \[\], list2 = \[0\]
**Output:** \[0\]
**Constraints:**
* The number of nodes in both lists is in the range `[0, 50]`.
* `-100 <= Node.val <= 100`
* Both `list1` and `list2` are sorted in **non-decreasing** order. | Solve the following problem in java, and explain your solution. |
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