What is the time complexity of Remove Linked List Elements?

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

What data structures are used to solve Remove Linked List Elements?

This problem uses Linked List, Recursion. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Remove Linked List Elements?

Using a dummy node simplifies handling the removal of the head node. Without a dummy node, removing the head requires special handling, making the code more complex. Iterative traversal of the linked list is efficient and straightforward for this problem. Recursion could be used but adds unnecessary overhead and potential stack overflow issues for very large lists.

Which companies ask LeetCode 203. Remove Linked List Elements in interviews?

This problem is frequently asked at Arista Networks.

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Remove Linked List Elements - LeetCode 203 Solution

Q: How to solve Remove Linked List Elements?

The provided solutions use an iterative approach with a dummy node. A dummy node is inserted at the beginning of the list. This simplifies the logic for removing nodes, especially the head node. The algorithm iterates through the list, and if a node's value matches the target value, it's bypassed by updating the `next` pointer of the previous node to skip the current node. Otherwise, the iterator moves to the next node. Finally, the `next` pointer of the dummy node is returned as the new head of the modified list.

Remove Linked List Elements - Complete Solution Guide

Remove Linked List Elements is LeetCode problem 203, 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 the head of a linked list and an integer val , remove all the nodes of the linked list that has Node.val == val , and return the new head . Example 1: Input: head = [1,2,6,3,4,5,6], val = 6 Output: [1,2,3,4,5] Example 2: Input: head = [], val = 1 Output: [] Example 3: Input: head = [7,7,7,7], val = 7 Output: [] Constraints: The number of nodes in the list is in the range [0, 10 4 ] . 1 <= Node.val <= 50 0 <= val <= 50

Detailed Explanation

The problem asks you to remove all nodes from a singly linked list that contain a specific integer value. The input is the head of the linked list and the integer value to remove. The output is the head of the modified linked list with all nodes containing the specified value removed. The linked list can be empty, and the input value might not be present in the list. The constraints specify a limit on the number of nodes and the range of values.

Solution Approach

The provided solutions use an iterative approach with a dummy node. A dummy node is inserted at the beginning of the list. This simplifies the logic for removing nodes, especially the head node. The algorithm iterates through the list, and if a node's value matches the target value, it's bypassed by updating the `next` pointer of the previous node to skip the current node. Otherwise, the iterator moves to the next node. Finally, the `next` pointer of the dummy node is returned as the new head of the modified list.

Step-by-Step Algorithm

Create a dummy node and point its `next` pointer to the head of the input linked list.
Initialize a current pointer to the dummy node.
Iterate through the list using the `current` pointer until the end of the list is reached (`current.next` is null).
If the value of the node pointed to by `current.next` is equal to the target value, update `current.next` to skip the node by setting it to `current.next.next`.
If the value does not match, move the `current` pointer to the next node (`current = current.next`).
Return the `next` pointer of the dummy node (this is the new head of the modified list).

Key Insights

Using a dummy node simplifies handling the removal of the head node. Without a dummy node, removing the head requires special handling, making the code more complex.
Iterative traversal of the linked list is efficient and straightforward for this problem. Recursion could be used but adds unnecessary overhead and potential stack overflow issues for very large lists.
The solution needs to correctly handle cases where the value to be removed is present at the head of the list or multiple times within the list.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Linked List, Recursion.

Companies

Asked at: Arista Networks.

Remove Linked List Elements | LeetCode 203

Easy

Linked List Recursion

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

Solution Code

Available in:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode removeElements(ListNode head, int val) {
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode current = dummy;

        while (current.next != null) {
            if (current.next.val == val) {
                current.next = current.next.next;
            } else {
                current = current.next;
            }
        }

        return dummy.next;
    }
}

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Problem Info

Description

Problem 203: Remove Linked List Elements - Detailed explanation and solution approaches.

Topics

Linked List Recursion

Companies

Arista Networks

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:203

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

6 steps

1
Create a dummy node and point its `next` pointer to the head of the input linked list.
2
Initialize a current pointer to the dummy node.
3
Iterate through the list using the `current` pointer until the end of the list is reached (`current.next` is null).
4
If the value of the node pointed to by `current.next` is equal to the target value, update `current.next` to skip the node by setting it to `current.next.next`.
5
If the value does not match, move the `current` pointer to the next node (`current = current.next`).
6
Return the `next` pointer of the dummy node (this is the new head of the modified list).

💡

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 algorithm iterates through the linked list once. The number of iterations is directly proportional to the number of nodes (n). No extra data structures are used beyond a few pointers; the space used remains constant regardless of the input size.

📖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

Recursive Approach

A recursive solution can traverse the list and recursively call the function for the next node, removing nodes with the target value as it goes.

Time: O(n)Space: O(n) - due to the recursive call stack

Trade-offs: The recursive approach is less efficient in terms of space complexity due to the potential for stack overflow with very large linked lists. The iterative approach is generally preferred for its better space efficiency.

✨

Key Insights

1Using a dummy node simplifies handling the removal of the head node. Without a dummy node, removing the head requires special handling, making the code more complex.
2Iterative traversal of the linked list is efficient and straightforward for this problem. Recursion could be used but adds unnecessary overhead and potential stack overflow issues for very large lists.
3The solution needs to correctly handle cases where the value to be removed is present at the head of the list or multiple times within the list.

📚

Key Terms Explained

For Students

Recursion

A function that calls itself to solve smaller versions of the same problem.

Linked List

Elements connected via pointers. Like a chain where each link knows the next.

Edge Cases

Empty list: The solution correctly handles an empty list by returning `null` (or `None` in Python).
All nodes to be removed: The solution correctly handles the case where all nodes in the list have the target value, resulting in an empty list being returned.
Target value not present: The solution correctly handles the case where the target value is not found in the list, leaving the list unchanged.

⚠️

Common Mistakes to Avoid

✗Forgetting to handle the case where the node to be removed is the head node. This can lead to memory leaks or incorrect results.

✗Incorrectly updating the `next` pointers, leading to a broken or incorrect linked list structure after removal.

✅

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?

💼Data cleanup in databases or systems where linked lists are used to represent data. Removing outdated or invalid entries.

🏦Filtering data streams. Imagine a stream of sensor readings where you want to ignore readings below a certain threshold. A linked list could be used to represent the stream, and this algorithm could filter out unwanted data.

💡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

The dummy node technique prevents edge cases. For cycle detection, Floyd's algorithm is the standard. Always handle empty list and single node cases.

🎯 Interview Tip

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

Losing reference to nodes when modifying pointers. Use temporary variables to save references.

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 24, 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

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

@CodeNinjaJul 13

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

@TechLeadJul 6

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 29

❓

Frequently Asked Questions

How does the fast/slow pointer technique work?

Fast pointer moves 2 steps while slow moves 1 step. When fast reaches the end, slow is at middle. If there's a cycle, they will meet. This technique is used for: finding middle node, detecting cycles, finding cycle start point, and checking palindromes.

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 Linked List Elements"?

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

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