What is the time complexity of Remove Nth Node From End of List?

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

What data structures are used to solve Remove Nth Node From End of List?

This problem uses Linked List, Two Pointers. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Remove Nth Node From End of List?

Insight 1: We need to find the nth node from the end without knowing the length of the list beforehand. Insight 2: Using two pointers with a fixed gap can help identify the node to be removed in a single pass.

Which companies ask LeetCode 19. Remove Nth Node From End of List in interviews?

This problem is frequently asked at Adobe, Amazon, Apple, Bloomberg, Citrix and 6 more companies.

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Remove Nth Node From End of List - LeetCode 19 Solution

Q: How to solve Remove Nth Node From End of List?

The solution uses the 'two-pointer' technique (also known as 'fast and slow pointers'). A 'fast' pointer moves 'n' steps ahead of the 'slow' pointer. Then, both pointers move forward simultaneously until the 'fast' pointer reaches the end of the list. At this point, the 'slow' pointer will be pointing to the node *before* the nth node from the end, allowing us to easily remove the nth node. A dummy node is used at the beginning of the list to handle the cases where the head of the list needs to be removed.

Remove Nth Node From End of List - Complete Solution Guide

Remove Nth Node From End of List is LeetCode problem 19, a Medium 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, remove the n th node from the end of the list and return its head. Example 1: Input: head = [1,2,3,4,5], n = 2 Output: [1,2,3,5] Example 2: Input: head = [1], n = 1 Output: [] Example 3: Input: head = [1,2], n = 1 Output: [1] Constraints: The number of nodes in the list is sz . 1 <= sz <= 30 0 <= Node.val <= 100 1 <= n <= sz Follow up: Could you do this in one pass?

Detailed Explanation

The problem requires us to remove the nth node from the end of a given singly linked list and return the modified list's head. The input is the head of the linked list and an integer 'n'. The output is the head of the updated linked list with the nth node from the end removed. The constraint is that 'n' will always be valid, meaning it will be within the bounds of the list's size. We need to accomplish this efficiently and ideally in one pass.

Solution Approach

The solution uses the 'two-pointer' technique (also known as 'fast and slow pointers'). A 'fast' pointer moves 'n' steps ahead of the 'slow' pointer. Then, both pointers move forward simultaneously until the 'fast' pointer reaches the end of the list. At this point, the 'slow' pointer will be pointing to the node *before* the nth node from the end, allowing us to easily remove the nth node. A dummy node is used at the beginning of the list to handle the cases where the head of the list needs to be removed.

Step-by-Step Algorithm

Step 1: Create a dummy node and set its 'next' pointer to the head of the linked list. This simplifies removing the head node if necessary.
Step 2: Initialize two pointers, 'slow' and 'fast', both pointing to the dummy node.
Step 3: Move the 'fast' pointer 'n' steps forward. Now, the 'fast' pointer is 'n' nodes ahead of the 'slow' pointer.
Step 4: Move both 'slow' and 'fast' pointers one step at a time until the 'fast' pointer reaches the end of the list (fast.next is null).
Step 5: At this point, the 'slow' pointer is pointing to the node immediately before the nth node from the end.
Step 6: Remove the nth node from the end by updating the 'next' pointer of the 'slow' node: `slow.next = slow.next.next`.
Step 7: Return the 'next' pointer of the dummy node, which is the head of the modified linked list.

Key Insights

Insight 1: We need to find the nth node from the end without knowing the length of the list beforehand.
Insight 2: Using two pointers with a fixed gap can help identify the node to be removed in a single pass.
Insight 3: Handling the edge case where the head itself needs to be removed is crucial. A dummy node simplifies this.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Linked List, Two Pointers.

Companies

Asked at: Adobe, Amazon, Apple, Bloomberg, Citrix, Meta, Microsoft, Nvidia, Qualcomm, Uber, Yahoo.

Remove Nth Node From End of List | LeetCode 19

Medium

Linked List Two Pointers

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 removeNthFromEnd(ListNode head, int n) {
        ListNode dummy = new ListNode(0, head);
        ListNode slow = dummy;
        ListNode fast = dummy;

        for (int i = 0; i < n; i++) {
            fast = fast.next;
        }

        while (fast.next != null) {
            slow = slow.next;
            fast = fast.next;
        }

        slow.next = slow.next.next;
        return dummy.next;
    }
}

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Problem Info

Description

Problem 19: Remove Nth Node From End of List - Detailed explanation and solution approaches.

Topics

Linked List Two Pointers

Companies

AdobeAmazonAppleBloombergCitrixMetaMicrosoftNvidia

Stats

Difficulty:Medium

Languages:4

Companies:11

Problem #:19

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

7 steps

1
Step 1: Create a dummy node and set its 'next' pointer to the head of the linked list. This simplifies removing the head node if necessary.
2
Step 2: Initialize two pointers, 'slow' and 'fast', both pointing to the dummy node.
3
Step 3: Move the 'fast' pointer 'n' steps forward. Now, the 'fast' pointer is 'n' nodes ahead of the 'slow' pointer.
4
Step 4: Move both 'slow' and 'fast' pointers one step at a time until the 'fast' pointer reaches the end of the list (fast.next is null).
5
Step 5: At this point, the 'slow' pointer is pointing to the node immediately before the nth node from the end.
6
Step 6: Remove the nth node from the end by updating the 'next' pointer of the 'slow' node: `slow.next = slow.next.next`.
7
Step 7: Return the 'next' pointer of the dummy node, which is the head of the modified linked 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 time complexity is linear because we visit each node in the list at most once. The space complexity is constant because we use a fixed number of variables irrespective of the input list 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

Calculate length first

First, traverse the entire list to calculate its length. Then, traverse the list again to reach the (length - n)th node and remove the (length - n + 1)th node.

Time: O(2n) which simplifies to O(n)Space: O(1)

Trade-offs: This approach requires two passes through the linked list, which is less efficient than the two-pointer approach, which only requires one pass. It's conceptually simpler to understand initially but less efficient. No dummy node needed.

✨

Key Insights

1Insight 1: We need to find the nth node from the end without knowing the length of the list beforehand.
2Insight 2: Using two pointers with a fixed gap can help identify the node to be removed in a single pass.
3Insight 3: Handling the edge case where the head itself needs to be removed is crucial. A dummy node simplifies this.

📚

Key Terms Explained

For Students

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Linked List

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

Edge Cases

Edge case 1: Empty list (head is null). Although the constraints state that the list size is between 1 and 30, it's good to consider. The code will not handle this case gracefully and would likely throw a null pointer exception during dummy node creation or when dereferencing head. However, the problem statement assumes a non-empty list.
Edge case 2: n is equal to the length of the list. This means we need to remove the head node. The dummy node is essential for correctly handling this case because the 'slow' pointer will be at the dummy node, and `slow.next = slow.next.next` will correctly remove the original head.
Edge case 3: n is 1. This means removing the last element. The algorithm handles this case correctly because slow will point to the second to last element and its next pointer will be set to null.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where the head of the list needs to be removed. Forgetting the dummy node will cause issues in this scenario.

✗Common mistake 2: Incorrectly calculating the distance between the two pointers. Ensure the 'fast' pointer moves 'n' steps ahead initially, not 'n-1' steps.

✗Common mistake 3: Null pointer exceptions. When moving the 'fast' pointer, check if it has reached the end of the list prematurely (before completing 'n' steps). This could happen if 'n' is greater than the length of the list (contrary to problem constraints).

✅

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: Managing a queue of tasks where you need to remove a task that is 'n' steps from the end of the queue.

🏦Real-world scenario 2: Implementing an undo/redo functionality in a text editor or image editing software, where you need to remove a specific action from the history of actions.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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 20, 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

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

@TechLeadJul 13

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

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 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.

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.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 Nth Node From End of List"?

For this medium-level Linked List 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.

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