Sort Array By Parity

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Given an array nums, return an array where all even numbers come first, followed by all odd numbers. The relative order within each group does not matter.

Example 1
Input: [3,1,2,4] → Output: [2,4,3,1]
Example 2
Input: [0] → Output: [0]

Solution Approach

Two pointers: lo starts at 0, hi at end. When lo is odd and hi is even, swap them.

  • lo=0, hi=n-1.
  • While lo
  • If lo
Java
class Solution {
    public int[] sortArrayByParity(int[] nums) {
        int lo = 0, hi = nums.length - 1;
        while (lo < hi) {
            while (lo < hi && nums[lo] % 2 == 0) lo++;
            while (lo < hi && nums[hi] % 2 == 1) hi--;
            if (lo < hi) {
                int tmp = nums[lo]; nums[lo] = nums[hi]; nums[hi] = tmp;
                lo++; hi--;
            }
        }
        return nums;
    }
}
  • Time Complexity: O(N)
  • Space Complexity: O(1)