LeetCode Study Note | Problem 283: Move Zeroes

🧩 Problem Description

Given an integer array nums, move all 0’s to the end of it while maintaining the relative order of the non-zero elements.

Example:

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

Constraints:

• Do this in-place without making a copy of the array.

• Minimize the total number of operations.

💡 Solution: Two Pointers

🌱 Core Idea

Use two pointers to iterate through the array:

• right pointer: scans through the array to find non-zero elements.

• left pointer: tracks the position where the next non-zero element should be placed.

🧠 Strategy

1. Traverse the array with right.

2. When nums[right] != 0:

   • Swap nums[right] with nums[left].

   • Move left one step to the right.

3. Continue until the end of the array.

This approach ensures all non-zero elements are moved to the front in order, and the leftover positions are filled with 0s.

🧪 Step-by-Step Example

Given nums = [0, 1, 0, 3, 12], let’s walk through:

Initial:   [0, 1, 0, 3, 12]
left = 0, right = 0 → nums[0] is 0 → do nothing
right = 1 → nums[1] = 1 ≠ 0 → swap nums[0] and nums[1]
→ [1, 0, 0, 3, 12], left++

right = 2 → nums[2] = 0 → do nothing
right = 3 → nums[3] = 3 → swap nums[1] and nums[3]
→ [1, 3, 0, 0, 12], left++

right = 4 → nums[4] = 12 → swap nums[2] and nums[4]
→ [1, 3, 12, 0, 0], left++

✅ Final result: [1, 3, 12, 0, 0]

🧾 Code (Java)

class Solution {
    public void moveZeroes(int[] nums) {
        int n = nums.length, left = 0, right = 0;
        while (right < n) {
            if (nums[right] != 0) {
                swap(nums, left, right);
                left++;
            }
            right++;
        }
    }

    private void swap(int[] nums, int i, int j) {
        if (i == j) return; // Optional: skip unnecessary swaps
        int temp = nums[i];
        nums[i] = nums[j];
        nums[j] = temp;
    }
}