make this as markdown"### **Problem: Move Zeroes**

**Goal**: Given an input list of integers **`nums`**, move all 0 elements to the end of the list while maintaining the relative order of the non-zero elements.

**Definition of a Zero**: Any element equal to 0.
**Element Separation**: Elements in **`nums`** are separated by at least one position.
**Objective**: Modify the input list **`nums`** in-place to move all zeros to the end, ensuring that the order of non-zero elements remains unchanged.

**Input Format**: A list of integers **`nums`**.

**Output Format**: The modified list **`nums`** with all zeros moved to the end.

### **Example**

For an input list:

```
nums = [0, 1, 0, 3, 12]

```

The expected result would be:

```
[1, 3, 12, 0, 0]

```

### **Approach**

You plan to solve this problem using a two-pointer technique to achieve an efficient in-place solution.

### **Understanding the Constraints:**

- **In-Place Modification**: The list must be modified without using extra space for another list.
- **Maintaining Order**: The relative order of non-zero elements should remain unchanged.
- **Efficiency**: Aim for an O(n) time complexity.

### **Initial Thought Process:**

**Two Pointers (i and j)**:

- **Pointer `i`**: Tracks the position where the next non-zero element should be placed.
- **Pointer `j`**: Iterates through the list to find non-zero elements.

**Logic**:

- Iterate through the list with pointer **`j`**.
- When a non-zero element is found at **`nums[j]`**, swap it with **`nums[i]`**.
- Increment both pointers accordingly.

**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

```

### **Explanation:**

1. Initialize **`i`** to 0.
2. Iterate through the list with **`j`**.
3. When a non-zero is found, swap it with the element at index **`i`**.
4. Increment **`i`** to point to the next position for a non-zero element.

### **Key Points**

- **Two-Pointer Technique**: Efficiently track and place non-zero elements.
- **In-Place Modification**: Swapping elements without using additional space.
- **Maintaining Order**: Ensuring that the relative order of non-zero elements remains unchanged.
- **Time Complexity**: Achieves O(n) by processing each element exactly once.
- **Space Complexity**: O(1), as it uses only constant extra space.

### **Feedback on Pseudocode**

Your pseudocode is almost correct but has a minor issue in the loop variable. Here’s the feedback:

- **Loop Variable Correction**: In your initial pseudocode, you used **`for i in range(n):`**, which incorrectly uses **`i`** as the loop variable. It should be **`for j in range(n):`** to iterate with pointer **`j`**.
- **Return Statement**: Since the modification is in-place, there's no need to return **`nums`**.

**Finalized 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

```

### **Manual Execution Example**

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

**Initial State**: **`nums = [0, 1, 0, 3, 12]`**, **`i = 0`**

- **Iteration 1 (j = 0)**:
    - **`nums[j] = 0`** (zero), no swap.
    - **`i`** remains 0.
- **Iteration 2 (j = 1)**:
    - **`nums[j] = 1`** (non-zero).
    - Swap **`nums[i]`** and **`nums[j]`** → **`nums = [1, 0, 0, 3, 12]`**.
    - Increment **`i`** to 1.
- **Iteration 3 (j = 2)**:
    - **`nums[j] = 0`** (zero), no swap.
    - **`i`** remains 1.
- **Iteration 4 (j = 3)**:
    - **`nums[j] = 3`** (non-zero).
    - Swap **`nums[i]`** and **`nums[j]`** → **`nums = [1, 3, 0, 0, 12]`**.
    - Increment **`i`** to 2.
- **Iteration 5 (j = 4)**:
    - **`nums[j] = 12`** (non-zero).
    - Swap **`nums[i]`** and **`nums[j]`** → **`nums = [1, 3, 12, 0, 0]`**.
    - Increment **`i`** to 3.

**Final State**: **`nums = [1, 3, 12, 0, 0]`**
All zeros have been moved to the end while maintaining the order of non-zero elements.

### **Final Solution**

Below is the final Python implementation of the Move Zeroes problem using the two-pointer technique:

```python
# 283. Move Zeroes.py

from typing import List

class Solution:
    def moveZeroes(self, nums: List[int]) -> None:
        """
        Move all zeros in 'nums' to the end while maintaining the order of non-zero elements.
        This is done in-place without making a copy of the array.
        """
        n = len(nums)
        i = 0  # Pointer for the position of the next non-zero element
        for j in range(n):
            if nums[j] != 0:
                nums[i], nums[j] = nums[j], nums[i]
                i += 1
        # No return statement needed as the list is modified in-place

```

### **How to Use**

- **Instantiate the Solution Class**:
    
    ```python
    solution = Solution()
    
    ```
    
- **Call the `moveZeroes` Method**:
    
    ```python
    nums = [0, 1, 0, 3, 12]
    solution.moveZeroes(nums)
    print(nums)  # Output: [1, 3, 12, 0, 0]
    
    ```
    

### **Summary**

Your approach effectively uses the two-pointer technique to solve the Move Zeroes problem with optimal time and space complexity. By carefully tracking the position for the next non-zero element and iterating through the list, all zeros are moved to the end in a single pass without disrupting the order of non-zero elements."