Shift All Zeros to the End of the Array

Easy10–15 min

Topics

ArrayTwo Pointer

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Given an array nums, return the array with every zero moved to the end, while every non-zero value keeps its original relative order. Try to do it by rearranging values in a single pass rather than building a second array — a two-pointer swap gets you there in O(1) extra space.

Test Case 1:

Input:nums = [0, 5, 0, 3, 9, 0]

Output:[5, 3, 9, 0, 0, 0]

Explanation:The non-zero values keep their original order; the zeros all land at the end.

Test Case 2:

Input:nums = [4, 2, 7]

Output:[4, 2, 7]

Explanation:No zeros to move, so nothing changes.

Test Case 3:

Input:nums = [0, 0, 0]

Output:[0, 0, 0]

Explanation:All zeros — moving them to the end leaves the array as it was.

Constraints

◆1 ≤ nums.length ≤ 10⁵
◆-10⁹ ≤ nums[i] ≤ 10⁹

🚀

Try the Dry Run

Don't just read the solution — watch it execute, one step at a time.

🧪Try your own test case

nums (array) — 1-10 integers

1	class Solution {
2	public int[] moveZeroesToEnd(int[] nums) {
3	int[] result = nums.clone();
4	int insertPos = 0;
5	for (int i = 0; i < result.length; i++) {
6	if (result[i] != 0) {
7	int tmp = result[insertPos];
8	result[insertPos] = result[i];
9	result[i] = tmp;
10	insertPos++;
11	}
12	}
13	return result;
14	}
15	}
16

↑insertPos

Variables

insertPos0

INITIALIZE

Copy nums into result, and set insertPos to 0 — that's where the next non-zero value will land.

Step 1 / 8

Approach & Solutions

Brute Force — Two Separate Arrays

Brute

Make one pass to collect every non-zero value into a fresh array, then append as many zeros as were left out. Easy to follow, but it needs a second array the same size as the input just to hold the answer, when the values could be rearranged directly.

TimeO(n)

SpaceO(n)

Java

Try it out

class Solution {
    public int[] moveZeroesToEnd(int[] nums) {
        int[] result = new int[nums.length];
        int idx = 0;
        for (int num : nums) {
            if (num != 0) result[idx++] = num;
        }
        while (idx < nums.length) result[idx++] = 0;
        return result;
    }
}

Optimal — Two-Pointer Swap

Optimal

Keep a pointer, insertPos, marking where the next non-zero value belongs. Scan the array once — whenever you find a non-zero value, swap it into position insertPos and advance that pointer. Every zero naturally gets pushed later in the array as swaps happen around it, with no second array needed.