Introduction

In this blog post, we will explore how to solve the LeetCode problem "922" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given an array of integers nums, half of the integers in nums are odd, and the other half are even. Sort the array so that whenever nums[i] is odd, i is odd, and whenever nums[i] is even, i is even. Return any answer array that satisfies this condition. Example 1: Input: nums = [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. Example 2: Input: nums = [2,3] Output: [2,3] Constraints: 2 <= nums.length <= 2 * 104 nums.length is even. Half of the integers in nums are even. 0 <= nums[i] <= 1000 Follow Up: Could you solve it in-place?

Explanation

Here's an efficient solution to the problem, along with explanations:

• Two Pointers: Use two pointers, one for even indices and one for odd indices.

• Iterate and Swap: Iterate through the array. If an element at an even index is odd, find an odd element at an odd index and swap them. Similarly, if an element at an odd index is even, find an even element at an even index, and swap them.

• Time and Space Complexity: O(n) time complexity, O(1) space complexity (in-place).

Code

    def sort_array_by_parity_ii(nums):
    """
    Sorts an array such that nums[i] is odd if i is odd and nums[i] is even if i is even.

    Args:
        nums: A list of integers.

    Returns:
        A list of integers sorted according to the parity condition.
    """
    n = len(nums)
    even_ptr = 0
    odd_ptr = 1

    while even_ptr < n and odd_ptr < n:
        if nums[even_ptr] % 2 == 0:
            even_ptr += 2
        elif nums[odd_ptr] % 2 != 0:
            odd_ptr += 2
        else:
            nums[even_ptr], nums[odd_ptr] = nums[odd_ptr], nums[even_ptr]
            even_ptr += 2
            odd_ptr += 2

    return nums