What is the time complexity of Reverse Nodes in k-Group?

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

What data structures are used to solve Reverse Nodes in k-Group?

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

What are the key insights for solving Reverse Nodes in k-Group?

Insight 1: The core operation is reversing a sublist of k nodes. This can be done iteratively. Insight 2: A dummy node simplifies handling the head of the linked list and the connections between reversed groups.

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Reverse Nodes in k-Group - LeetCode 25 Solution

Q: Which companies ask LeetCode 25. Reverse Nodes in k-Group in interviews?

This problem is frequently asked at Adobe, Amazon, Apple, Arista Networks, Autodesk and 23 more companies.

Reverse Nodes in k-Group - Complete Solution Guide

Reverse Nodes in k-Group is LeetCode problem 25, a Hard 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, reverse the nodes of the list k at a time, and return the modified list . k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes, in the end, should remain as it is. You may not alter the values in the list's nodes, only nodes themselves may be changed. Example 1: Input: head = [1,2,3,4,5], k = 2 Output: [2,1,4,3,5] Example 2: Input: head = [1,2,3,4,5], k = 3 Output: [3

Detailed Explanation

The problem asks us to reverse the nodes of a singly linked list in groups of `k`. Given the head of a linked list and an integer `k`, we need to reverse the order of nodes within each group of `k` nodes. If the number of nodes is not a multiple of `k`, the remaining nodes at the end should remain in their original order. We are constrained to only modify the node pointers and not the node values themselves. For example, if the linked list is `1 -> 2 -> 3 -> 4 -> 5` and `k = 2`, the output should be `2 -> 1 -> 4 -> 3 -> 5`. If `k = 3`, the output should be `3 -> 2 -> 1 -> 4 -> 5`.

Solution Approach

The solution uses an iterative approach to reverse the linked list in k-sized groups. A dummy node is created to simplify the process, especially when dealing with the head of the list. The algorithm iterates through the linked list, identifying groups of `k` nodes. For each group, it reverses the order of the nodes. If there are fewer than `k` nodes remaining, the algorithm stops without reversing them. The reversed groups are then connected to form the final reversed list.

Step-by-Step Algorithm

Step 1: Create a dummy node and point it to the head of the linked list. This simplifies handling the head of the list.
Step 2: Initialize `group_prev` to the dummy node. This pointer will keep track of the node preceding each group to be reversed.
Step 3: In a `while` loop, check if there are at least `k` nodes remaining. This is done by traversing `k` nodes from `group_prev`. If there are fewer than `k` nodes, return the `dummy.next` as the reversed list is complete.
Step 4: Identify the last node (`kth`) in the current group to be reversed, and also find the node after the kth node called `group_next`. This `group_next` node is used later to connect with the reversed group.
Step 5: Reverse the nodes within the current group of `k` nodes using the iterative reversal technique. Use `prev`, `curr`, and `tmp` pointers to perform the reversal.
Step 6: Connect the reversed group to the rest of the linked list. This involves updating the `next` pointer of `group_prev` to point to the `kth` (last node of reversed sublist) and updating the pointer `group_prev` to the first node of the reversed group, originally pointed by `group_prev.next` before reversal.
Step 7: Repeat steps 3-6 until there are fewer than `k` nodes remaining.

Key Insights

Insight 1: The core operation is reversing a sublist of k nodes. This can be done iteratively.
Insight 2: A dummy node simplifies handling the head of the linked list and the connections between reversed groups.
Insight 3: We need to check if there are at least `k` nodes remaining before attempting to reverse a group. Otherwise, the remaining nodes should be left as is.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Linked List, Recursion.

Companies

Asked at: Adobe, Amazon, Apple, Arista Networks, Autodesk, Bloomberg, Capital One, Commvault, DE Shaw, Infosys, MakeMyTrip, MathWorks, Meta, Microsoft, Nutanix, PornHub, Qualcomm, Sigmoid, Snowflake, Tesla, TikTok, VMware, Visa, Walmart Labs, Yahoo, Zeta, Zopsmart, josh technology.

Reverse Nodes in k-Group | LeetCode 25

Hard

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 reverseKGroup(ListNode head, int k) {
        if (head == null || k == 1) {
            return head;
        }

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

        while (true) {
            ListNode kth = group_prev;
            for (int i = 0; i < k; i++) {
                kth = kth.next;
                if (kth == null) {
                    return dummy.next;
                }
            }
            
            ListNode group_next = kth.next;
            
            ListNode prev = group_next;
            ListNode curr = group_prev.next;
            while (curr != group_next) {
                ListNode tmp = curr.next;
                curr.next = prev;
                prev = curr;
                curr = tmp;
            }
            
            ListNode tmp = group_prev.next;
            group_prev.next = kth;
            group_prev = tmp;
        }
    }
}

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Problem Info

Description

Problem 25: Reverse Nodes in k-Group - Detailed explanation and solution approaches.

Topics

Linked List Recursion

Companies

AdobeAmazonAppleArista NetworksAutodeskBloombergCapital OneCommvault

Stats

Difficulty:Hard

Languages:4

Companies:28

Problem #:25

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 point it to the head of the linked list. This simplifies handling the head of the list.
2
Step 2: Initialize `group_prev` to the dummy node. This pointer will keep track of the node preceding each group to be reversed.
3
Step 3: In a `while` loop, check if there are at least `k` nodes remaining. This is done by traversing `k` nodes from `group_prev`. If there are fewer than `k` nodes, return the `dummy.next` as the reversed list is complete.
4
Step 4: Identify the last node (`kth`) in the current group to be reversed, and also find the node after the kth node called `group_next`. This `group_next` node is used later to connect with the reversed group.
5
Step 5: Reverse the nodes within the current group of `k` nodes using the iterative reversal technique. Use `prev`, `curr`, and `tmp` pointers to perform the reversal.
6
Step 6: Connect the reversed group to the rest of the linked list. This involves updating the `next` pointer of `group_prev` to point to the `kth` (last node of reversed sublist) and updating the pointer `group_prev` to the first node of the reversed group, originally pointed by `group_prev.next` before reversal.
7
Step 7: Repeat steps 3-6 until there are fewer than `k` nodes remaining.

💡

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 O(n) because we iterate through the linked list at most twice for each node. The space complexity is O(1) because we only use a constant amount of extra memory, irrespective of the size of the input list. The algorithm manipulates the list in-place using a few pointers. The space used does not scale with the size of the input.

📖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 also be used to solve this problem. It involves reversing the first k nodes and then recursively calling the function on the remaining list.

Time: O(n)Space: O(n/k) in the worst case where k = 1, leading to O(n) stack space. In best case if k = n then it will O(1).

Trade-offs: The recursive approach might be more concise to write, but it uses stack space, which can be a disadvantage for very large lists. The iterative approach is generally preferred due to its constant space complexity.

✨

Key Insights

1Insight 1: The core operation is reversing a sublist of k nodes. This can be done iteratively.
2Insight 2: A dummy node simplifies handling the head of the linked list and the connections between reversed groups.
3Insight 3: We need to check if there are at least `k` nodes remaining before attempting to reverse a group. Otherwise, the remaining nodes should be left as is.

📚

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

Edge case 1: Empty linked list. The solution correctly handles an empty list by returning it directly since `if not head` at the start.
Edge case 2: k = 1. If k is 1, no reversal is needed, so the original list is returned. The condition `k == 1` at the beginning handles this case.
Edge case 3: k is greater than the number of nodes. The solution checks if there are at least `k` nodes before reversing. If not, it returns the remaining list as is, preventing errors.
Edge case 4: k equals the number of nodes. The entire list is reversed.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case where the number of nodes is not a multiple of k. Failing to properly handle remaining nodes can lead to incorrect results.

✗Common mistake 2: Incorrectly connecting the reversed group to the rest of the list. It is crucial to update the `next` pointers correctly to ensure that the reversed groups are properly linked.

✗Common mistake 3: Not considering the edge case of an empty list or k=1. This can lead to unexpected errors or incorrect output.

✅

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: This problem can be related to data processing scenarios where data needs to be processed in chunks and the order within each chunk needs to be reversed. For example, reversing the order of packets in a network stream for specific applications.

🏦Real-world scenario 2: It can also be applied in data structure manipulations in databases or file systems where specific segments of data need to be reorganized or reversed.

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

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

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

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 26, 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

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

@DevExpert92Jul 13

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

@AlgoMasterJul 6

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

@InterviewProJun 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.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 "Reverse Nodes in k-Group"?

For this hard-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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