---

### Problem: Reverse Words in a String

**Goal**: Given an input string `s`, reverse the order of the words.

- **Definition of a Word**: A "word" here is defined as a sequence of non-space characters.
- **Word Separation**: Words in `s` are separated by at least one space.

**Objective**: Return a string where the words are in reverse order and are joined by a single space.

**Input Format**: A single string `s`.
**Output Format**: A single string with words in reversed order.

### Example

For an input string:

```
s = "this sky is blue"

```

The expected result would be:

```
"blue is sky this"

```

---

### Approach

You plan to solve this problem using a functional programming approach.

1. **Step 1**: Implement a helper function `helper(s: str) -> str` to reverse a given string.
    - **Explanation**:
        - This helper function takes a string `s` as input and reverses its characters. Using a two-pointer technique, it swaps characters from the start (`left`) and end (`right`) until they meet in the middle.
    
    ```python
    def helper(s: str) -> str:
        n = len(s)
        left = 0
        right = n - 1
        while left < right:
            s[left], s[right] = s[right], s[left]
            left += 1
            right -= 1
        return s
    
    ```
    
2. **Step 2**: Apply the helper function to each word in the input string.
    - **Explanation**:
        - Define a main function `solution(words: str) -> str`.
        - First, split the input string `words` into a list of words (`word_list`).
        - Use the `map` function to apply `helper()` to each word, then collect the results into a list `mapped_s`.
        - Finally, return the reversed list of words joined by a single space.
    
    ```python
    def solution(words: str) -> str:
        word_list = words.split()
        mapped_s = list(map(helper, word_list))
        return " ".join(mapped_s)
    
    ```
    

---

### Key Points

- The `helper` function reverses each word.
- The `solution` function reverses the order of words in the input string.
- The final result joins these words with a single space.

---

Let me know if you need any further clarifications!

![image.png](https://prod-files-secure.s3.us-west-2.amazonaws.com/c26cbd8c-4ae9-4ab6-b595-99bb653f9d83/ddf047a9-f01e-4cfe-9cbb-b006dac8fd98/image.png)

Q- bahi honii 3ana problem “esmo Move Zeros” bahii fe this problem we have an inter list “nums” , 

bahii honii lzemik we move “O’s” to the end of it while maintining the relative order of the non-zero elements 

Q- what is the input & output format ? 

![image.png](https://prod-files-secure.s3.us-west-2.amazonaws.com/c26cbd8c-4ae9-4ab6-b595-99bb653f9d83/c83823d4-3d42-4f96-9435-eff957484173/image.png)

input : List[int] 

output : List[int] 

Q- is 

If there is any constraint.

yes they is a constrain in this problem because we need to mutated in place 

Q- next : 

bahii in this problem we need to use two pointer for usre but the two question 

where did we need to initiate them 

and what is the logic to update them ? 

bahii we need to aplly this logic 

[0,1,0,3,12] 

if found out a zeor 

i = 0 —> 0 

so  we need to sifte the i+1 —> n element to the left and insert i at the end okii 

Q- this my first apporch but i do khnow if it’s fessible using twopointers i and j 

q- Without giving me the full solution, just give me feedback about this view.

Your initial approach of shifting elements left whenever you encounter a zero isn't the most efficient way to solve this using two pointers. It leads to repeated shifting and has a time complexity of O(n^2) in the worst case (imagine an array full of zeros except for the last element). Two pointers can provide an O(n) solution.

Q- okii we can elemimniate the first approch ? 

Q- what next ? 

bhaii what about a pointer that track the “zeros” and a pointer that track the non zero  okii 

they will scan the full array at the end to first iteration 

zero_pointer will point to last occurent zero 

non_zero_pointer will point to the last occueent zero ? 

the if zero_pointer < non_zero_pointer : 

then swap them ? 

q- 

The feedback about this approach. Can you give me a manual execution of this process using any example?

Your idea of tracking zeros and non-zeros separately is getting closer, but the current implementation won't work correctly. The issue lies in how the pointers are used and updated. Swapping just the last occurrences won't necessarily move all zeros to the end while preserving the order of non-zero elements.

Q- okii the serach now is getting closer okii ? 

—> let’ try to think of logic that 

how to use the two pointer better here ? 

did at high level 

Tracking the zero and non-zero elements with 2-pointer is at high level a good approach here.

yes or non 

Yes, at a high level, the concept of using two pointers to manage zero and non-zero elements is a valid and efficient strategy for this problem. The crucial part that needs refinement is the *specific logic* of how those pointers are used and updated during the array traversal. The previous explanations highlighted the inefficiency of simply swapping the last occurrences.

Q- bhii is the two pointer start from the zero index i mean top left ? 

Yes, in the efficient two-pointer approach for the Move Zeroes problem, both pointers typically start at the beginning of the array (index 0). One pointer (i) tracks the position where the next non-zero element should be placed, and the other pointer (j) scans through the array looking for non-zero values. They are not tracking zeros and non-zeros separately, but rather working together to place non-zeros at the front and leave the zeros for the end.

Q- bahii honii : 

we just need to use “i” to track the non-zero element where it need to be placed 

and “j” search for non zero elements to place them corrrectly okii 

for exmple 

[0,1,0,3,12] 

here 

i—>0 so , [0] so this the first element where the non-zero element will be placed on 

j—>0 point to 0 zero element so we upate it j+1 

k=1

i→ 0 : still point to 0 

j—> 1 : now point to 1 , a non zero element okii 

so we swap it with (nums[i] , nums[j]) 

update i to be j 

and also updtate i 

Q- 

And now did my approach getting closer to the right direction?

Yes, your understanding is indeed getting closer to an efficient solution! Using two pointers like this, where one tracks the position for the next non-zero element (`i`) and the other pointer (`j`) scans for non-zero elements, is a sound strategy.

Here’s how this refined approach works step-by-step:

1. **`i` starts at the beginning (index 0)** and represents the "next position" where a non-zero element should go.
2. **`j` also starts at index 0** and iterates through each element in the list.
3. If `nums[j]` is non-zero, **swap** `nums[i]` and `nums[j]`, and **increment `i`** to the next position.
4. If `nums[j]` is zero, simply move to the next element by incrementing `j`.

By the end of the iteration, all non-zero elements will be placed at the start of the list, and zeros will naturally follow.

This approach has a **time complexity of O(n)**, since each element is processed once, and a **space complexity of O(1)**, as it operates in-place.

### Manual Execution Example:

Using your example `nums = [0, 1, 0, 3, 12]`:

- Initial state: `nums = [0, 1, 0, 3, 12]`, `i = 0`, `j = 0`
1. `j = 0`: `nums[j] = 0` (no swap), increment `j`.
2. `j = 1`: `nums[j] = 1` (non-zero), swap `nums[i]` and `nums[j]` → `nums = [1, 0, 0, 3, 12]`, increment `i`.
3. `j = 2`: `nums[j] = 0` (no swap), increment `j`.
4. `j = 3`: `nums[j] = 3` (non-zero), swap `nums[i]` and `nums[j]` → `nums = [1, 3, 0, 0, 12]`, increment `i`.
5. `j = 4`: `nums[j] = 12` (non-zero), swap `nums[i]` and `nums[j]` → `nums = [1, 3, 12, 0, 0]`, increment `i`.

Final output: `nums = [1, 3, 12, 0, 0]`, with all zeros moved to the end while preserving the order of non-zero elements.

Your refined approach now matches the optimal method for solving this problem!

Q- so here we have build the right mental model here : 

Q- now we only need to check out how to implement this : 

Q- so let’s start the first iteration ? 

```python
PseudoCode : Functon (nums:List[int])->int: 
					// get the length of the array
					n=len(nums) 
					i=0
					for j in range(n):
							if nums[j]!=0 and j!=i:
									nums[i],nums[j]=nums[j],nums[i] 
									i=j
					return nums 
```

Q

Here, this is my first iteration of the pseudocode. Try to give me feedback without giving me the full solution, ok?

Your pseudocode is close to the final solution, but there are a few adjustments to make:

1. **Update `i` correctly:** In your code, `i = j` after the swap may not work as expected because `i` is meant to track the next position for a non-zero element, not necessarily `j`’s position. After swapping, simply increment `i` rather than setting it equal to `j`.
2. **Condition Simplification:** The condition `j != i` isn't necessary for swapping here. Instead, let `i` and `j` function independently—`j` iterates through each element, while `i` only increments when a non-zero element is placed.
3. **Minor Corrections:** Instead of returning `nums` in the pseudocode, we typically don’t return anything for in-place modifications.

Here’s a slightly refined version of your pseudocode:

```python
Function moveZeroes(nums: List[int]) -> None:
    n = len(nums)
    i = 0
    for j in range(n):
        if nums[j] != 0:
            nums[i], nums[j] = nums[j], nums[i]
            i += 1

```

### Key Points:

- **`i` only increments** after a non-zero element is placed, which keeps all zeros naturally at the end.
- This approach ensures **O(n)** time complexity and **O(1)** space complexity.

This refined logic aligns with the optimal two-pointer method for solving the problem.

Q- now lets try to get the final solution in python ? 

```python
class Solution:
    def moveZeroes(self, nums: List[int]) -> None:
				n =len(nums)
				i=0
				for i in range(n):
						if nums[j]!=0:
							nums[i],nums[j]=nums[j],nums[i] 
							i+=1
				return nums 
```

Q create a python file called “

# [**283. Move Zeroes](https://leetcode.com/problems/move-zeroes/).py**

Q- and inset this code 

```python
class Solution:
    def moveZeroes(self, nums: List[int]) -> None:
        n = len(nums)
        i = 0
        for j in range(n):  # Corrected loop variable
            if nums[j] != 0:
                nums[i], nums[j] = nums[j], nums[i]
                i += 1
        # No return statement needed as we modify in place
```

Q- now do nit remove any latter and 

These are the style of my notes, just formatted in a good way. the full artical 

Q- use rthe same style okii and do no change it