What is the time complexity of Reverse Nodes in Even Length Groups?

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

What data structures are used to solve Reverse Nodes in Even Length Groups?

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

What are the key insights for solving Reverse Nodes in Even Length Groups?

Insight 1: Identifying the start and end of each group is crucial. Since the group lengths are sequential natural numbers, we can maintain a counter to track the current group's length. Insight 2: The core operation involves reversing a sublist within the linked list. A standard linked list reversal algorithm needs to be applied when a group's length is even.

Which companies ask LeetCode 2074. Reverse Nodes in Even Length Groups in interviews?

This problem is frequently asked at Zopsmart.

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Reverse Nodes in Even Length Groups - LeetCode 2074 Solution

Reverse Nodes in Even Length Groups - Complete Solution Guide

Reverse Nodes in Even Length Groups is LeetCode problem 2074, 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

You are given the head of a linked list. The nodes in the linked list are sequentially assigned to non-empty groups whose lengths form the sequence of the natural numbers ( 1, 2, 3, 4, ... ). The length of a group is the number of nodes assigned to it. In other words, The 1 st node is assigned to the first group. The 2 nd and the 3 rd nodes are assigned to the second group. The 4 th , 5 th , and 6 th nodes are assigned to the third group, and so on. Note that the length of the last group may be

Detailed Explanation

The problem requires you to modify a given singly-linked list by reversing nodes within specific groups. The linked list nodes are divided into groups of increasing size (1, 2, 3, 4, ...). If a group has an even number of nodes, you must reverse the order of nodes within that group. The goal is to return the head of the modified linked list after performing these reversals.

Solution Approach

The provided solution uses an iterative approach. It traverses the linked list, identifying groups based on an increasing group length. For each group, it checks if the length is even. If so, it reverses the nodes within the group using a standard iterative linked list reversal. After reversing, it carefully re-links the reversed group into the main linked list. If the group length is odd, no reversal is done, and the algorithm moves to the next group.

Step-by-Step Algorithm

Step 1: Create a dummy node. This simplifies handling the head of the linked list and edge cases.
Step 2: Initialize `prev_group_tail` to the dummy node. This pointer will track the tail of the previously processed group, allowing us to link the current group after potential reversal.
Step 3: Initialize `group_len` to 1, representing the length of the first group.
Step 4: Iterate while `prev_group_tail.next` is not null. This ensures we process all nodes in the list.
Step 5: In each iteration, identify the `group_head` (the first node of the current group) and `group_tail` (the last node of the current group). Traverse `group_len` nodes, or until the end of the list, whichever comes first.
Step 6: Store the `next_group_head` (the node after the current group) to assist in relinking.
Step 7: Check if the `count` (actual length of the current group) is even.
Step 8: If `count` is even, reverse the nodes from `group_head` to `group_tail`. Use a standard iterative list reversal.
Step 9: After reversing, update pointers: `prev_group_tail.next` points to the new head of the reversed group (the original `group_tail`), and `prev_group_tail` is updated to the original `group_head` (which is now the tail of the reversed group).
Step 10: If `count` is odd, simply move `prev_group_tail` to `group_tail` without reversing.
Step 11: Increment `group_len` for the next group.
Step 12: After the loop finishes, return `dummy.next`, which points to the head of the modified linked list.

Key Insights

Insight 1: Identifying the start and end of each group is crucial. Since the group lengths are sequential natural numbers, we can maintain a counter to track the current group's length.
Insight 2: The core operation involves reversing a sublist within the linked list. A standard linked list reversal algorithm needs to be applied when a group's length is even.
Insight 3: Correctly linking the reversed sublist with the rest of the list is essential. Pay close attention to updating the `next` pointers of the nodes surrounding the reversed section.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Linked List.

Companies

Asked at: Zopsmart.

Reverse Nodes in Even Length Groups | LeetCode 2074

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 reverseEvenLengthGroups(ListNode head) {
        if (head == null) {
            return null;
        }

        ListNode dummy = new ListNode(0, head);
        ListNode prev_group_tail = dummy;
        int group_len = 1;

        while (prev_group_tail.next != null) {
            ListNode group_head = prev_group_tail.next;
            ListNode group_tail = group_head;
            int count = 1;

            int i = 1;
            while (i < group_len && group_tail.next != null) {
                group_tail = group_tail.next;
                count += 1;
                i += 1;
            }

            ListNode next_group_head = group_tail.next;

            if (count % 2 == 0) {
                // Reverse the current group
                ListNode curr = group_head;
                ListNode prev = next_group_head;

                for (int j = 0; j < count; j++) {
                    ListNode next_node = curr.next;
                    curr.next = prev;
                    prev = curr;
                    curr = next_node;
                }

                // After reversal, the original group_tail is the new head
                // and the original group_head is the new tail.
                prev_group_tail.next = group_tail;
                prev_group_tail = group_head;
            } else {
                prev_group_tail = group_tail;
            }

            group_len += 1;
        }

        return dummy.next;
    }
}

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Problem Info

Description

Problem 2074: Reverse Nodes in Even Length Groups - Detailed explanation and solution approaches.

Topics

Linked List

Companies

Zopsmart

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2074

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

12 steps

1
Step 1: Create a dummy node. This simplifies handling the head of the linked list and edge cases.
2
Step 2: Initialize `prev_group_tail` to the dummy node. This pointer will track the tail of the previously processed group, allowing us to link the current group after potential reversal.
3
Step 3: Initialize `group_len` to 1, representing the length of the first group.
4
Step 4: Iterate while `prev_group_tail.next` is not null. This ensures we process all nodes in the list.
5
Step 5: In each iteration, identify the `group_head` (the first node of the current group) and `group_tail` (the last node of the current group). Traverse `group_len` nodes, or until the end of the list, whichever comes first.
6
Step 6: Store the `next_group_head` (the node after the current group) to assist in relinking.
7
Step 7: Check if the `count` (actual length of the current group) is even.
8
Step 8: If `count` is even, reverse the nodes from `group_head` to `group_tail`. Use a standard iterative list reversal.
9
Step 9: After reversing, update pointers: `prev_group_tail.next` points to the new head of the reversed group (the original `group_tail`), and `prev_group_tail` is updated to the original `group_head` (which is now the tail of the reversed group).
10
Step 10: If `count` is odd, simply move `prev_group_tail` to `group_tail` without reversing.
11
Step 11: Increment `group_len` for the next group.
12
Step 12: After the loop finishes, return `dummy.next`, which points to 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 O(n) because each node is visited a constant number of times (at most twice: once during group identification and potentially once during reversal). The space complexity is O(1) because only a few pointers are used, and their number doesn't depend on 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 for Reversal

Instead of iteratively reversing a sublist, one could implement a recursive reversal function. This would achieve the same result but might be slightly less efficient due to function call overhead.

Time: O(n)Space: O(k) where k is the maximum group size. This is due to the recursive call stack.

Trade-offs: The recursive approach may be more readable for some but might consume more memory, particularly with large linked lists.

✨

Key Insights

1Insight 1: Identifying the start and end of each group is crucial. Since the group lengths are sequential natural numbers, we can maintain a counter to track the current group's length.
2Insight 2: The core operation involves reversing a sublist within the linked list. A standard linked list reversal algorithm needs to be applied when a group's length is even.
3Insight 3: Correctly linking the reversed sublist with the rest of the list is essential. Pay close attention to updating the `next` pointers of the nodes surrounding the reversed section.

📚

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. The code handles this case at the beginning by returning `None` (or `null` in Java/C++).
Edge case 2: List with only one node. The loop condition `prev_group_tail.next` will be false after the first group (length 1), so no reversal occurs, and the list remains unchanged, which is the correct behavior.
Edge case 3: Last group has fewer nodes than expected. The inner `while` loop ensures that the traversal stops when it reaches the end of the list, even if the current group's length is not yet fully satisfied.
Edge case 4: Last group has an even length. The reversal logic handles this case correctly, re-linking the reversed group to the rest of the list.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly relinking the reversed sublist. Forgetting to update `prev_group_tail.next` or updating it with the wrong node will break the list structure.

✗Common mistake 2: Not handling the case where the last group has fewer nodes than the current group length. Failing to check for the end of the list in the inner loop can lead to accessing invalid memory.

✗Common mistake 3: Modifying the `next` pointer of a node before storing a reference to the next node. This can lead to losing the remaining part of the list.

✗Common mistake 4: Not handling the dummy node correctly which could cause issues at head of list.

✅

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: Implementing undo/redo functionality. Linked lists are used to maintain a history of actions, and reversing a sublist could be part of reverting to a previous state.

🏦Real-world scenario 2: Media players or playlist management systems. Certain functionalities might require reversing a section of the playlist dynamically.

🎮Real-world scenario 3: Text editors or word processors can use linked lists to manage text. Reversing even length groups could simulate some text transformations or special effects.

💡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: September 3, 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 14

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

@DevExpert92Jul 7

❓

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 Nodes in Even Length Groups"?

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.

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