09 Sept 2026•Go / Python / TypeScript•Easy

Sort Array By Parity

Reorder an array so all even values appear before all odd values.

View original LeetCode question

Partition in place with pointers from both ends, swapping an odd left value with an even right value.

complexity

O(n) time and O(1) extra space.

solution files

Go sort-array-by-parity/solution.go
Python sort-array-by-parity/solution.py
TypeScript sort-array-by-parity/solution.ts

Solution files

Go•sort-array-by-parity/solution.go

package main

func sortArrayByParity(nums []int) []int {
	left, right := 0, len(nums)-1
	for left < right {
		for left < right && nums[left]%2 == 0 {
			left++
		}
		for left < right && nums[right]%2 != 0 {
			right--
		}
		nums[left], nums[right] = nums[right], nums[left]
	}
	return nums
}

Python•sort-array-by-parity/solution.py

class Solution:
    def sortArrayByParity(self, nums: list[int]) -> list[int]:
        left, right = class="syntax-number">0, len(nums) - class="syntax-number">1
        while left < right:
            while left < right and nums[left] % class="syntax-number">2 == class="syntax-number">0: left += class="syntax-number">1
            while left < right and nums[right] % class="syntax-number">2: right -= class="syntax-number">1
            nums[left], nums[right] = nums[right], nums[left]
        return nums

TypeScript•sort-array-by-parity/solution.ts

function sortArrayByParity(nums: number[]): number[] {
  let left = class="syntax-number">0;
  let right = nums.length - class="syntax-number">1;
  while (left < right) { while (left < right && nums[left] % class="syntax-number">2 === class="syntax-number">0) left += class="syntax-number">1; while (left < right && nums[right] % class="syntax-number">2 !== class="syntax-number">0) right -= class="syntax-number">1; [nums[left], nums[right]] = [nums[right], nums[left]]; }
  return nums;
}