Leetcode offers a rich set of problems that sharpen a developer’s problem-solving skills. One such interesting problem is “Sort Array By Parity II,” which pushes a coder to think about array manipulation in creative ways. This article is dedicated to providing an in-depth understanding of the problem, its underlying principles, Python solutions, and a detailed complexity analysis.

## Table of Contents:

- Understanding the Problem Statement
- Problem Analysis
- Two-Pointer Technique
- Python Implementation
- Complexity Analysis
- Alternate Approaches
- Conclusion

## 1. Understanding the Problem Statement

### Problem:

Given an array `A`

of non-negative integers, half of the integers in `A`

are odd, and half of the integers are even. Sort the array so that whenever `A[i]`

is odd, `i`

is odd; and whenever `A[i]`

is even, `i`

is even. You may return any answer array that satisfies this condition.

**Example**:

```
Input: [4,2,5,7]
Output: [4,5,2,7]
Explanation: [4,7,2,5], [2,5,4,7], [2,7,4,5] would also have been accepted.
```

## 2. Problem Analysis

A key insight from the problem is that there’s a clear distinction between the positions of even and odd numbers. Our primary objective is to ensure that even indices have even numbers and odd indices have odd numbers.

## 3. Two-Pointer Technique

A smart way to tackle this problem is by using the two-pointer technique. In this case, we can use:

`even`

: a pointer starting at 0, moving in increments of 2.`odd`

: a pointer starting at 1, moving in increments of 2.

We iterate until our `even`

pointer finds an odd number, and our `odd`

pointer finds an even number. Once this condition is met, we swap these two elements.

## 4. Python Implementation

Below is the Python code for the described approach:

```
def sortArrayByParityII(A):
even, odd, n = 0, 1, len(A)
while even < n and odd < n:
while even < n and A[even] % 2 == 0:
even += 2
while odd < n and A[odd] % 2 == 1:
odd += 2
if even < n and odd < n:
A[even], A[odd] = A[odd], A[even]
return A
```

## 5. Complexity Analysis

**Time Complexity**: O(n)

The algorithm goes through the list a maximum of two times, leading to a linear time complexity.

**Space Complexity**: O(1)

No additional space proportional to the input size is used, leading to a constant space complexity.

## 6. Alternate Approaches

**Two new arrays**: Create two new arrays – one for even numbers and one for odd numbers. Then, merge these two arrays to form the result, picking alternately from the even array and the odd array.**In-place sorting**: Use a custom comparator function to sort the array. While this might work for certain languages/platforms, it isn’t recommended due to the increased time complexity of the sorting operation.

## 7. Conclusion

The “Sort Array By Parity II” problem offers a great example of the benefits of the two-pointer technique, especially when dealing with array manipulations. Solving problems like these help in refining the approach towards array-based questions, promoting a better understanding of how to harness pointers efficiently.