LeetCode 905: Sort Array By Parity Solution

Master LeetCode problem 905 (Sort Array By Parity), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

905. Sort Array By Parity

Easy

Array Two Pointers Sorting

Problem Explanation

Explanation

To solve this problem, we can use a two-pointer approach where we maintain two pointers left and right. The left pointer starts at the beginning of the array and moves to the right, while the right pointer starts at the end of the array and moves to the left. We swap elements at these pointers if they are at the wrong place (even number on the right side or odd number on the left side). The process continues until the left pointer is greater than or equal to the right pointer.

Time complexity: O(n) where n is the number of elements in the array.
Space complexity: O(1) constant extra space is used.

Solution Code

class Solution {
    public int[] sortArrayByParity(int[] nums) {
        int left = 0;
        int right = nums.length - 1;
        
        while (left < right) {
            if (nums[left] % 2 != 0 && nums[right] % 2 == 0) {
                int temp = nums[left];
                nums[left] = nums[right];
                nums[right] = temp;
            }
            
            if (nums[left] % 2 == 0) {
                left++;
            }
            
            if (nums[right] % 2 != 0) {
                right--;
            }
        }
        
        return nums;
    }
}

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