Coding problem · Easy · Basics

Move Zeroes to the End

Easy Algorithms & data structures

Return nums with all 0s moved to the end, keeping the relative order of the non-zero numbers.

Example 1

Input: nums = [0,1,0,3,12]
Output: [1,3,12,0,0]

Example 2

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

Hints

Hint 1

Collect the non-zero numbers first, then add as many zeros as you skipped.

Hint 2

In place: a 'write' index marks where the next non-zero number goes.

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Solution and explanation

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Time O(n) · Space O(1)

Copy the non-zero values in order, then pad with zeros. The in-place version uses two pointers: write for the next non-zero slot, read scanning every element.

Reference solution

function moveZeroes(nums) {
  const nonZero = nums.filter((n) => n !== 0);
  return nonZero.concat(new Array(nums.length - nonZero.length).fill(0));
}

function moveZeroes(nums: number[]): number[] {
  const nonZero = nums.filter((n) => n !== 0);
  return nonZero.concat(new Array<number>(nums.length - nonZero.length).fill(0));
}

def move_zeroes(nums: list[int]) -> list[int]:
    non_zero = [n for n in nums if n != 0]
    return non_zero + [0] * (len(nums) - len(non_zero))

<?php

function moveZeroes(array $nums): array
{
    $nonZero = array_values(array_filter($nums, fn ($n) => $n !== 0));
    return array_merge($nonZero, array_fill(0, count($nums) - count($nonZero), 0));
}

class Solution {
    public int[] moveZeroes(int[] nums) {
        int[] out = new int[nums.length];
        int write = 0;
        for (int n : nums) if (n != 0) out[write++] = n;
        return out;
    }
}

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