LeetCode Day32 Dynamic Programming Part 5
Tags: leetcode, java, algorithms
Published: 2024-07-12 04:56:13

# 518. Coin Change II
You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money.

Return the number of combinations that make up that amount. If that amount of money cannot be made up by any combination of the coins, return 0.

You may assume that you have an infinite number of each kind of coin.

The answer is guaranteed to fit into a signed 32-bit integer.

 

Example 1:

Input: amount = 5, coins = [1,2,5]
Output: 4
Explanation: there are four ways to make up the amount:
5=5
5=2+2+1
5=2+1+1+1
5=1+1+1+1+1
Example 2:

Input: amount = 3, coins = [2]
Output: 0
Explanation: the amount of 3 cannot be made up just with coins of 2.
Example 3:

Input: amount = 10, coins = [10]
Output: 1
 

Constraints:

1 <= coins.length <= 300
1 <= coins[i] <= 5000
All the values of coins are unique.
0 <= amount <= 5000
[Original Page](https://leetcode.com/problems/coin-change-ii/description/)

```
    public int change(int amount, int[] coins) {
        int[][] dp = new int[coins.length+1][amount+1];
        for(int i=0; i<=coins.length; i++){
            dp[i][0] = 1;
        }
        for(int i=1; i<= coins.length; i++){
            for(int j=1; j<=amount; j++){
                if(j<coins[i-1]){
                    dp[i][j] = dp[i-1][j];
                }else{
                    dp[i][j] = dp[i-1][j] + dp[i][j-coins[i-1]];
                }
                
            }
        }
                    // Arrays.stream(dp).map(Arrays::toString).forEach(System.out::println);
        return dp[coins.length][amount];
    }
```



```
    public int change(int amount, int[] coins) {
        int[] dp = new int[amount+1];
        dp[0] = 1;
        for(int i=0; i< coins.length; i++){
            for(int j=coins[i]; j<=amount; j++){
                dp[j] = dp[j] + dp[j-coins[i]];
            }
            System.out.println(Arrays.toString(dp));
        }
        return dp[amount];
    }
```


# 377. Combination Sum IV
Given an array of distinct integers nums and a target integer target, return the number of possible combinations that add up to target.

The test cases are generated so that the answer can fit in a 32-bit integer.

 

Example 1:

Input: nums = [1,2,3], target = 4
Output: 7
Explanation:
The possible combination ways are:
(1, 1, 1, 1)
(1, 1, 2)
(1, 2, 1)
(1, 3)
(2, 1, 1)
(2, 2)
(3, 1)
Note that different sequences are counted as different combinations.
Example 2:

Input: nums = [9], target = 3
Output: 0
 

Constraints:

1 <= nums.length <= 200
1 <= nums[i] <= 1000
All the elements of nums are unique.
1 <= target <= 1000
 

Follow up: What if negative numbers are allowed in the given array? How does it change the problem? What limitation we need to add to the question to allow negative numbers?


```
    public int combinationSum4(int[] nums, int target) {
        int[] dp = new int[target+1];
        dp[0] = 1;
        for(int i=1; i<=target; i++){
            for(int j=0; j<nums.length; j++){
                if(nums[j] <= i){
                    dp[i] = dp[i] + dp[i-nums[j]];
                }
            }
            // System.out.println(Arrays.toString(dp));
        }
        return dp[target];
    }
```