Move All Zeros to the End of an Array in JavaScript (2026 Complete Interview Guide)
Last updated for 2026 | JavaScript Coding Interview | Frontend & Angular Developers
If you are preparing for JavaScript interviews in 2026, Angular frontend roles, or improving your core coding fundamentals, this problem is one of the most asked and high-demand interview questions:
Move all
0s to the end of an array while maintaining the order of non-zero elements.
This question frequently appears in:
- JavaScript coding interviews
- Angular / React frontend interviews
- Online coding platforms (HackerRank, LeetCode, CodeSignal)
- Machine coding & logic rounds
🔍 Problem Statement
You are given an array of numbers. Your task is to move all zero (0) values to the end of the array while preserving the relative order of non-zero elements.
Example
Input: [0, 1, 3, 0, 5, 0]
Output: [1, 3, 5, 0, 0, 0]
📘 What Is an Array in JavaScript?
An array in JavaScript is a data structure used to store multiple values in a single variable.
const arr = [1, 2, 3, 4];
Key Properties of Arrays
- Arrays are zero-indexed
- They are mutable (can be modified)
- Can store mixed data types
- Support built-in methods like push, splice, filter, map
Arrays are heavily used in frontend frameworks like Angular and React, especially when rendering lists, handling API data, and state management.
🧠 Approach 1: Using splice() (Naive / Less Optimal)
🔧 What is splice() in JavaScript?
splice() is a built-in JavaScript array method used to add, remove, or replace elements in an array.
array.splice(startIndex, deleteCount)
Example
const arr = [1, 2, 3, 4];
arr.splice(1, 1); // removes element at index 1
console.log(arr); // [1, 3, 4]
Code Implementation (Using splice)
const newArray = [0, 1, 3, 0, 5, 0];
let count = 0;
for (let i = 0; i < newArray.length; i++) {
if (newArray[i] === 0) {
newArray.splice(i, 1); // remove zero
count++;
i--; // important to avoid skipping elements
}
}
for (let i = 0; i < count; i++) {
newArray.push(0);
}
console.log(newArray); // [1, 3, 5, 0, 0, 0]
❌ Why This Approach Is NOT Recommended
- splice() shifts array elements internally
- Causes O(n) work per removal
- Leads to O(n²) time complexity in worst case
- Risky if index adjustment is missed
⏱ Complexity
- Time Complexity: O(n²)
- Space Complexity: O(1)
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⭐ Approach 2: Two-Pointer / Overwrite Technique (BEST & INTERVIEW-READY)
This is the most efficient and professional solution, widely accepted in 2026 coding interviews.
🧩 Concept
Use one pointer to track where the next non-zero element should go
Copy non-zero elements forward
Fill remaining positions with zeros
Code Implementation
const newArray = [0, 1, 3, 0, 5, 0];
let index = 0;
// Step 1: Move non-zero elements forward
for (let i = 0; i < newArray.length; i++) {
if (newArray[i] !== 0) {
newArray[index] = newArray[i];
index++;
}
}
// Step 2: Fill remaining positions with zeros
for (let i = index; i < newArray.length; i++) {
newArray[i] = 0;
}
console.log(newArray); // [1, 3, 5, 0, 0, 0]
🔍 Dry Run Explanation
Input:
[0, 1, 3, 0, 5, 0]
After moving non-zeros:
[1, 3, 5, 0, 5, 0]
After filling zeros:
[1, 3, 5, 0, 0, 0]
✔ Order preserved
✔ In-place solution
✔ No extra memory
🏆 Complexity Comparison
| Approach | Time Complexity | Space Complexity | Interview Friendly |
|---|---|---|---|
| splice() | O(n²) | O(1) | ❌ No |
| Two-Pointer | O(n) | O(1) | ✅ YES |
🎯 Why Interviewers Prefer the Two-Pointer Approach
- Predictable performance
- No array resizing
- Clean and readable logic
- Safe for large datasets
Interview Tip (2026): Avoid using
splice()inside loops unless explicitly required.
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🏷️ Tags
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📌 Final Verdict
If your goal is to:
- Crack JavaScript / Angular interviews in 2026
- Write efficient and clean code
- Impress interviewers with optimal logic
👉 Always use the two-pointer overwrite technique.
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