11. Move Zeroes · DSA Practice

Move all 0s to the end of nums in place, keeping the relative order of the non-zero elements. Do not return anything.

Example 1

Input: nums = [4,0,0,5,6]

Output: [4,5,6,0,0]

Example 2

Input: nums = [0]

Output: [0]

Hint 1

Use a “write” pointer for where the next non-zero should go.

Two Pointers

Approach 1: Extra array

Copy non-zeros then pad zeros.

⏱ Time O(n)💾 Space O(n)

Approach 2: Read / write pointers ✅

Swap each non-zero into the write position.

⏱ Time O(n)💾 Space O(1)

Reference solution for Read / write pointers (time O(n), space O(1)). Each solution passes every test in all four languages.

class Solution:
    def moveZeroes(self, nums: List[int]) -> None:
        w = 0                                  # next position for a non-zero
        for r in range(len(nums)):
            if nums[r] != 0:
                nums[w], nums[r] = nums[r], nums[w]
                w += 1

class Solution {
    public void moveZeroes(int[] nums) {
        int w = 0;                             // next position for a non-zero
        for (int r = 0; r < nums.length; r++)
            if (nums[r] != 0) { int t = nums[w]; nums[w] = nums[r]; nums[r] = t; w++; }
    }
}

class Solution {
public:
    void moveZeroes(vector<int>& nums) {
        int w = 0;                             // next position for a non-zero
        for (int r = 0; r < (int)nums.size(); r++)
            if (nums[r] != 0) swap(nums[w++], nums[r]);
    }
};

function moveZeroes(nums) {
  let w = 0; // next position for a non-zero
  for (let r = 0; r < nums.length; r++)
    if (nums[r] !== 0) { [nums[w], nums[r]] = [nums[r], nums[w]]; w++; }
}

Try solving it yourself first. Interviewers care more about how you explain the approach than about the finished code.

This problem is explained in the lesson Two Pointers.