What is the time complexity of Remove Element?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Remove Element?

This problem uses Array, Two Pointers. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Remove Element?

**In-place compaction via a write pointer:** The core idea is to maintain a `k` pointer that tracks the *next available position* for an element that should be kept. This pointer only increments when a valid element is encountered and written, effectively compacting the array to the left. **Leveraging 'order may be changed' for simplicity:** Since the relative order of elements not equal to `val` doesn't need to be preserved, we don't need complex shifting. We can simply overwrite the 'removed' elements with valid ones found later in the array, making the single-pass copy operation straightforward.

Which companies ask LeetCode 27. Remove Element in interviews?

This problem is frequently asked at Accenture, Adobe, Amazon, Apple, Bloomberg and 5 more companies.

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Remove Element - LeetCode 27 Solution

Remove Element - Complete Solution Guide

Remove Element is LeetCode problem 27, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given an integer array nums and an integer val , remove all occurrences of val in nums in-place . The order of the elements may be changed. Then return the number of elements in nums which are not equal to val . Consider the number of elements in nums which are not equal to val be k , to get accepted, you need to do the following things: Change the array nums such that the first k elements of nums contain the elements which are not equal to val . The remaining elements of nums are not important

Detailed Explanation

This problem asks us to modify an integer array, `nums`, in-place. Our goal is to 'remove' all occurrences of a specific integer, `val`. The twist is that we're not actually shrinking the array's physical size. Instead, we need to shift all elements *not* equal to `val` to the beginning of the array. The specific requirement is that if there are `k` such elements, the first `k` positions of `nums` must contain them, and the values in `nums` beyond `k` don't matter. Finally, we must return this count `k`.

Solution Approach

The elegance of the provided solution lies in its efficient use of a single-pass, two-pointer approach, albeit a slightly specialized one. We essentially use one pointer, `i`, to traverse the entire original array from left to right, acting as our 'read' pointer. A second pointer, `k`, serves as our 'write' pointer. `k` only advances when we encounter an element (`nums[i]`) that is *not* equal to `val`. When such an element is found, we copy it to `nums[k]` and then increment `k`. If `nums[i]` *is* equal to `val`, we simply skip it. By doing this, `k` effectively keeps track of the next available position for an element that should be kept, ensuring all non-`val` elements are continuously compacted to the beginning of the array. The elements equal to `val` are implicitly 'removed' by being overwritten later by valid elements, or simply left in the irrelevant tail portion of the array.

Step-by-Step Algorithm

Step 1: Initialize a counter `k` to 0. This variable will keep track of the index where the next element different from `val` should be placed.
Step 2: Iterate through the input array `nums` using a loop (from index 0 to `nums.length -1`).
Step 3: Inside the loop, check if the current element `nums[i]` is not equal to `val`.
Step 4: If `nums[i]` is not equal to `val`, copy its value to `nums[k]`. Then, increment `k` to point to the next available position for elements different from `val`.
Step 5: After the loop completes, `k` will hold the number of elements in the array that are not equal to `val`. Return `k`.

Key Insights

**In-place compaction via a write pointer:** The core idea is to maintain a `k` pointer that tracks the *next available position* for an element that should be kept. This pointer only increments when a valid element is encountered and written, effectively compacting the array to the left.
**Leveraging 'order may be changed' for simplicity:** Since the relative order of elements not equal to `val` doesn't need to be preserved, we don't need complex shifting. We can simply overwrite the 'removed' elements with valid ones found later in the array, making the single-pass copy operation straightforward.
**Single-pass linear scan:** The solution processes each element of the array exactly once with the `i` pointer, and each kept element is written at most once by the `k` pointer. This results in an optimal O(N) time complexity, making it highly efficient for large inputs.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Two Pointers.

Companies

Asked at: Accenture, Adobe, Amazon, Apple, Bloomberg, Meta, Microsoft, Uber, Yahoo, Yandex.

Remove Element | LeetCode 27

Easy

Array Two Pointers

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int removeElement(int[] nums, int val) {
        int k = 0;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] != val) {
                nums[k] = nums[i];
                k++;
            }
        }
        return k;
    }
}

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Problem Info

Description

Problem 27: Remove Element - Detailed explanation and solution approaches.

Topics

Array Two Pointers

Companies

AccentureAdobeAmazonAppleBloombergMetaMicrosoftUber

Stats

Difficulty:Easy

Languages:4

Companies:10

Problem #:27

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

5 steps

1
Step 1: Initialize a counter `k` to 0. This variable will keep track of the index where the next element different from `val` should be placed.
2
Step 2: Iterate through the input array `nums` using a loop (from index 0 to `nums.length -1`).
3
Step 3: Inside the loop, check if the current element `nums[i]` is not equal to `val`.
4
Step 4: If `nums[i]` is not equal to `val`, copy its value to `nums[k]`. Then, increment `k` to point to the next available position for elements different from `val`.
5
Step 5: After the loop completes, `k` will hold the number of elements in the array that are not equal to `val`. Return `k`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The loop iterates through each element of the array exactly once. No additional data structures that scale with the input size are used. Therefore, both time and space complexity are linear and constant, respectively.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Creating a new array

Iterate through the original array and create a new array containing only the elements that are not equal to `val`. Then, copy the elements from the new array back into the original array (if required by problem statement).

Time: O(n)Space: O(n)

Trade-offs: This approach is less efficient in terms of space complexity compared to the two-pointer approach, especially for large arrays. The two-pointer method is preferred because it's in-place and more memory-efficient.

✨

Key Insights

1**In-place compaction via a write pointer:** The core idea is to maintain a `k` pointer that tracks the *next available position* for an element that should be kept. This pointer only increments when a valid element is encountered and written, effectively compacting the array to the left.
2**Leveraging 'order may be changed' for simplicity:** Since the relative order of elements not equal to `val` doesn't need to be preserved, we don't need complex shifting. We can simply overwrite the 'removed' elements with valid ones found later in the array, making the single-pass copy operation straightforward.
3**Single-pass linear scan:** The solution processes each element of the array exactly once with the `i` pointer, and each kept element is written at most once by the `k` pointer. This results in an optimal O(N) time complexity, making it highly efficient for large inputs.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Edge Cases

Edge case 1: Empty array ([]). The code handles this correctly; the loop won't execute, and `k` remains 0, which is the correct return value.
Edge case 2: Array where all elements are equal to `val`. The loop will run, but the `if` condition will always be false. `k` will remain 0, correctly indicating that no elements are different from `val`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the in-place requirement correctly. Some beginners might create a new array to store the results, violating the in-place constraint.

✗Common mistake 2: Incorrectly managing the index `k`. Off-by-one errors are common when incrementing or using the `k` index.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Data cleaning and preprocessing. Removing unwanted or invalid data points from a dataset before further processing.

🏦Real-world scenario 2: Filtering lists or arrays in applications. For example, removing duplicate entries or entries that meet a certain criteria from a database record.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 28, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I avoid infinite loops with two pointers?

Always ensure your loop condition guarantees termination (typically left < right or slow != fast). Move at least one pointer in each iteration, and handle edge cases where pointers might not move (like skipping duplicates). Draw out pointer movement on paper first.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Remove Element"?

For this easy-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether two pointers techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?