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Introduction

LeetCode 283: Move Zeroes asks you to reorder elements in an array by moving all zeros to the end while maintaining the relative order of the non-zero elements.

This is one of the best array manipulation and two-pointer problems you’ll encounter in interviews — clean, efficient, and elegant.

Problem Statement

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

• Do this in-place without making a copy of the array.
• Minimize the total number of operations.

Example

python

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

✅ Step-by-Step Solution

🧠 Intuition

We can use a two-pointer approach:

• One pointer (insert_pos) keeps track of where the next non-zero element should go.
• The other pointer (i) scans the array.

🔄 Step-by-Step Process

1. Initialize insert_pos = 0
2. Loop through the array:

• If nums[i] != 0:
• Place nums[i] at nums[insert_pos]
• Increment insert_pos

1. After loop ends, fill the rest of the array from insert_pos to end with 0

✅ Python Code

python

class Solution:
    def moveZeroes(self, nums: list[int]) -> None:
        insert_pos = 0
        for i in range(len(nums)):
            if nums[i] != 0:
                nums[insert_pos] = nums[i]
                insert_pos += 1

        for i in range(insert_pos, len(nums)):
            nums[i] = 0

This modifies the array in-place without extra space.

🧪 Dry Run Example

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

• i=0: 0 → skip
• i=1: 1 → place at insert_pos=0, insert_pos=1
• i=2: 0 → skip
• i=3: 3 → place at insert_pos=1, insert_pos=2
• i=4: 12 → place at insert_pos=2, insert_pos=3

Fill rest with 0s:

→ [1, 3, 12, 0, 0]

⏱️ Time and Space Complexity

MetricValueTime ComplexityO(n)Space ComplexityO(1)In-Place✅ Yes

🛠️ Alternative: Swap-Based (Slightly Different Behavior)

python

class Solution:
    def moveZeroes(self, nums: list[int]) -> None:
        last_non_zero = 0
        for i in range(len(nums)):
            if nums[i] != 0:
                nums[i], nums[last_non_zero] = nums[last_non_zero], nums[i]
                last_non_zero += 1

• This version swaps instead of assigning.
• Works well but may introduce more operations than necessary.

🔍 Edge Cases

• [0, 0, 0] → stays [0, 0, 0]
• [1, 2, 3] → no change
• [] → empty list → no-op

✅ Conclusion

LeetCode 283: Move Zeroes is a fantastic warm-up for interview-style array problems. It teaches:

• Two-pointer strategies
• In-place transformation
• Optimal movement minimization