Introduction

In this blog post, we will explore how to solve the LeetCode problem "905" 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 integer array nums, move all the even integers at the beginning of the array followed by all the odd integers. Return any array that satisfies this condition. Example 1: Input: nums = [3,1,2,4] Output: [2,4,3,1] Explanation: The outputs [4,2,3,1], [2,4,1,3], and [4,2,1,3] would also be accepted. Example 2: Input: nums = [0] Output: [0] Constraints: 1 <= nums.length <= 5000 0 <= nums[i] <= 5000

Explanation

Here's the solution to rearrange the array such that even numbers come before odd numbers:

• Two-Pointer Approach: Use two pointers, left and right, initialized at the beginning and end of the array, respectively.
• Move Pointers: Move the left pointer to the right until it points to an odd number and move the right pointer to the left until it points to an even number.

• Swap: If left is less than right, swap the elements at these pointers. Repeat until left and right cross each other.

• Time Complexity: O(n), Space Complexity: O(1)

Code

    def sort_array_by_parity(nums: list[int]) -> list[int]:
    """
    Given an integer array nums, move all the even integers at the beginning of the array
    followed by all the odd integers.
    Return any array that satisfies this condition.

    Example 1:
    Input: nums = [3,1,2,4]
    Output: [2,4,3,1]
    Explanation: The outputs [4,2,3,1], [2,4,1,3], and [4,2,1,3] would also be accepted.

    Example 2:
    Input: nums = [0]
    Output: [0]

    Constraints:
    1 <= nums.length <= 5000
    0 <= nums[i] <= 5000
    """
    left, right = 0, len(nums) - 1

    while left < right:
        if nums[left] % 2 == 0:
            left += 1
        elif nums[right] % 2 != 0:
            right -= 1
        else:
            nums[left], nums[right] = nums[right], nums[left]
            left += 1
            right -= 1

    return nums