Sort Array by Parity
(#905), Easy
Category: Array, Sorting, Two Pointer
Problem Statement
You're given an array of integers nums, and your task is to move all even numbers to the front of the array, followed by all the odd numbers. You can return any valid arrangement that satisfies this condition.
Examples
Input: [3, 1, 2, 4]
Output: [4, 2, 1, 3] or [2, 4, 1, 3]
Input: [0]
Output: [0]
Input: [1, 2, 3, 4]
Output: [2, 4, 3, 1]
Approach
Two-Pointer Approach
- Initialize two pointers: `left = 0` and `right = nums.length - 1`.
- While `left < right`:
- - If `nums[left]` is odd and `nums[right]` is even, swap them.
- - If `nums[left]` is even, increment `left`.
- - If `nums[right]` is odd, decrement `right`.
- This ensures all even numbers move to the beginning of the array.
var sortArrayByParity = function (nums) {
let left = 0;
let right = nums.length - 1;
while (left < right) {
if (nums[left] % 2 > nums[right] % 2) {
[nums[left], nums[right]] = [nums[right], nums[left]];
}
if (nums[left] % 2 === 0) left++;
if (nums[right] % 2 === 1) right--;
}
return nums;
};Complexity
Time: O(n) – Each element is checked at most once.
Space: O(1) – In-place swapping, no extra space used.