What is the time complexity of Reverse Linked List II?

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

What data structures are used to solve Reverse Linked List II?

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

What are the key insights for solving Reverse Linked List II?

Insight 1: Using a dummy node simplifies the process of handling the head node and avoids special case logic when 'left' is 1. Insight 2: Iterative reversal is efficient and avoids recursion's overhead.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Reverse Linked List II - LeetCode 92 Solution

Q: How to solve Reverse Linked List II?

The solution uses an iterative approach to reverse the specified portion of the linked list. A dummy node is used to handle the case where 'left' is 1. The algorithm first locates the node immediately before the 'left' position. Then, it iteratively reverses the nodes from 'left' to 'right'. Finally, it reconnects the reversed sublist with the rest of the linked list.

Q: Which companies ask LeetCode 92. Reverse Linked List II in interviews?

This problem is frequently asked at Adobe, Amazon, Apple, Arista Networks, Bloomberg and 11 more companies.

Reverse Linked List II - Complete Solution Guide

Reverse Linked List II is LeetCode problem 92, 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 singly linked list and two integers left and right where left <= right , reverse the nodes of the list from position left to position right , and return the reversed list . Example 1: Input: head = [1,2,3,4,5], left = 2, right = 4 Output: [1,4,3,2,5] Example 2: Input: head = [5], left = 1, right = 1 Output: [5] Constraints: The number of nodes in the list is n . 1 <= n <= 500 -500 <= Node.val <= 500 1 <= left <= right <= n Follow up: Could you do it in one pass?

Detailed Explanation

The problem requires reversing a specific portion of a singly linked list, defined by the 'left' and 'right' indices (1-indexed). Given the head of the linked list and the left and right boundaries, the goal is to reverse the nodes from the 'left' position up to and including the 'right' position. The reversed sublist should then be re-integrated into the original linked list. The constraints specify that the list size is between 1 and 500, node values between -500 and 500, and left/right are valid indices within the list's boundaries.

Solution Approach

The solution uses an iterative approach to reverse the specified portion of the linked list. A dummy node is used to handle the case where 'left' is 1. The algorithm first locates the node immediately before the 'left' position. Then, it iteratively reverses the nodes from 'left' to 'right'. Finally, it reconnects the reversed sublist with the rest of the linked list.

Step-by-Step Algorithm

Step 1: Create a dummy node pointing to the head of the linked list. This simplifies the logic when 'left' is 1.
Step 2: Traverse the linked list to find the node immediately before the 'left' position. This node is stored as 'pre_left'.
Step 3: Identify the starting node of the section to be reversed, which is 'pre_left.next', and store it as 'start_node'.
Step 4: Initialize 'prev_node' to null, 'current_node' to 'start_node'.
Step 5: Iterate 'right - left + 1' times to reverse the sublist: store the next node, reverse the 'next' pointer of the current node to point to the previous node, move 'prev_node' to the current node, and move 'current_node' to the next node.
Step 6: After reversing the sublist, reconnect it to the original list: set 'pre_left.next' to 'prev_node' (the new head of the reversed sublist) and set 'start_node.next' to 'current_node' (the node immediately after the reversed sublist).
Step 7: Return 'dummy.next', which is the head of the modified linked list.

Key Insights

Insight 1: Using a dummy node simplifies the process of handling the head node and avoids special case logic when 'left' is 1.
Insight 2: Iterative reversal is efficient and avoids recursion's overhead.
Insight 3: Keeping track of the node immediately before the 'left' position and the node at the 'left' position before reversal allows for easy reconnection of the reversed sublist.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Linked List.

Companies

Asked at: Adobe, Amazon, Apple, Arista Networks, Bloomberg, Disney, EPAM Systems, Meta, Microsoft, Nutanix, Nvidia, Revolut, TikTok, Uber, Yahoo, Zoho.

Reverse Linked List II | LeetCode 92

Medium

Linked List

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 reverseBetween(ListNode head, int left, int right) {
        if (head == null || left == right) {
            return head;
        }

        ListNode dummy = new ListNode(0, head);
        ListNode pre_left = dummy;

        for (int i = 0; i < left - 1; i++) {
            pre_left = pre_left.next;
        }
        
        ListNode start_node = pre_left.next;
        
        ListNode prev_node = null;
        ListNode current_node = start_node;
        
        for (int i = 0; i < right - left + 1; i++) {
            ListNode next_node = current_node.next;
            current_node.next = prev_node;
            prev_node = current_node;
            current_node = next_node;
        }
        
        pre_left.next = prev_node;
        start_node.next = current_node;
        
        return dummy.next;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 92: Reverse Linked List II - Detailed explanation and solution approaches.

Topics

Linked List

Companies

AdobeAmazonAppleArista NetworksBloombergDisneyEPAM SystemsMeta

Stats

Difficulty:Medium

Languages:4

Companies:16

Problem #:92

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 pointing to the head of the linked list. This simplifies the logic when 'left' is 1.
2
Step 2: Traverse the linked list to find the node immediately before the 'left' position. This node is stored as 'pre_left'.
3
Step 3: Identify the starting node of the section to be reversed, which is 'pre_left.next', and store it as 'start_node'.
4
Step 4: Initialize 'prev_node' to null, 'current_node' to 'start_node'.
5
Step 5: Iterate 'right - left + 1' times to reverse the sublist: store the next node, reverse the 'next' pointer of the current node to point to the previous node, move 'prev_node' to the current node, and move 'current_node' to the next node.
6
Step 6: After reversing the sublist, reconnect it to the original list: set 'pre_left.next' to 'prev_node' (the new head of the reversed sublist) and set 'start_node.next' to 'current_node' (the node immediately after the reversed sublist).
7
Step 7: Return 'dummy.next', 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 each node is visited at most a constant number of times. The space complexity is constant because only a few pointers are used.

📖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

Reversing the linked list section can be done recursively. However, it would generally be less efficient due to function call overhead.

Time: O(n)Space: O(n) in the worst case, due to call stack depth.

Trade-offs: While conceptually simpler for some, it's less efficient than the iterative approach in terms of space complexity due to recursion depth. Therefore, the iterative approach is preferred.

✨

Key Insights

1Insight 1: Using a dummy node simplifies the process of handling the head node and avoids special case logic when 'left' is 1.
2Insight 2: Iterative reversal is efficient and avoids recursion's overhead.
3Insight 3: Keeping track of the node immediately before the 'left' position and the node at the 'left' position before reversal allows for easy reconnection of the reversed sublist.

📚

Key Terms Explained

For Students

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). The code handles this by returning the head directly, as there's nothing to reverse.
Edge case 2: left == right. If left and right are equal, no reversal is needed, so the original list is returned.
Edge case 3: left = 1. When left is 1, the dummy node simplifies the logic of reattaching the reversed list to the beginning of the original list.
Edge case 4: right = n. If right is equal to number of nodes, the right most section of linked list has to be reversed

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to reconnect the reversed sublist to the rest of the linked list correctly, especially the tail of the reversed portion.

✗Common mistake 2: Not handling the case where 'left' is 1 properly, leading to incorrect results when reversing from the beginning of the list.

✗Common mistake 3: Incorrectly iterating to find pre_left and start_node, leading to reversing the wrong portion of linkedlist

✅

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: Undo functionality in text editors or other applications where a sequence of actions needs to be reversed. Operations can be stored as nodes in a linked list and reversed when the undo button is pressed.

🏦Real-world scenario 2: Network packet processing where certain segments of a data stream need to be reordered or reversed for specific protocols or security measures.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ 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: September 1, 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 13

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

@CodeNinjaJul 6

❓

Frequently Asked Questions

Why use a dummy node in linked list problems?

A dummy node simplifies edge cases where the head might change (insertion/deletion at head) or when building a new list. Initialize dummy.next = head, work with dummy.next throughout, and return dummy.next at the end to handle all cases uniformly.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Reverse Linked List II"?

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 standard techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?