What is the time complexity of Maximum Twin Sum of a Linked List?

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

What data structures are used to solve Maximum Twin Sum of a Linked List?

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

What are the key insights for solving Maximum Twin Sum of a Linked List?

Insight 1: The problem can be solved by finding the middle of the linked list and reversing the second half. Insight 2: Using the two-pointer technique (slow and fast pointers) is efficient for finding the middle node of the linked list.

Which companies ask LeetCode 2130. Maximum Twin Sum of a Linked List in interviews?

This problem is frequently asked at josh technology.

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Maximum Twin Sum of a Linked List - LeetCode 2130 Solution

Q: How to solve Maximum Twin Sum of a Linked List?

The solution involves three main steps: finding the middle of the linked list, reversing the second half of the linked list, and calculating the maximum twin sum by iterating through the first half and the reversed second half. First, use slow and fast pointers to find the middle. Then, reverse the linked list starting from the middle node. Finally, iterate from the head of the original linked list and the head of the reversed second half, calculating the twin sum at each step and updating the maximum twin sum as needed.

Maximum Twin Sum of a Linked List - Complete Solution Guide

Maximum Twin Sum of a Linked List is LeetCode problem 2130, 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

In a linked list of size n , where n is even , the i th node ( 0-indexed ) of the linked list is known as the twin of the (n-1-i) th node, if 0 <= i <= (n / 2) - 1 . For example, if n = 4 , then node 0 is the twin of node 3 , and node 1 is the twin of node 2 . These are the only nodes with twins for n = 4 . The twin sum is defined as the sum of a node and its twin. Given the head of a linked list with even length, return the maximum twin sum of the linked list . Example 1: Input: head = [5,4,2,1

Detailed Explanation

The problem asks us to find the maximum twin sum in a linked list. A twin sum is the sum of a node's value and the value of its 'twin' node. The twin of the i-th node (0-indexed) is the node at the (n-1-i)-th position, where 'n' is the number of nodes in the linked list. The linked list has an even number of nodes. The goal is to iterate through the linked list, calculate the twin sum for each node in the first half, and return the maximum twin sum found.

Solution Approach

The solution involves three main steps: finding the middle of the linked list, reversing the second half of the linked list, and calculating the maximum twin sum by iterating through the first half and the reversed second half. First, use slow and fast pointers to find the middle. Then, reverse the linked list starting from the middle node. Finally, iterate from the head of the original linked list and the head of the reversed second half, calculating the twin sum at each step and updating the maximum twin sum as needed.

Step-by-Step Algorithm

Step 1: Initialize two pointers, 'slow' and 'fast', both pointing to the head of the linked list.
Step 2: Move 'slow' one step at a time and 'fast' two steps at a time until 'fast' reaches the end of the list (or 'fast.next' is null). At this point, 'slow' will be pointing to the middle of the list.
Step 3: Reverse the second half of the linked list starting from 'slow'. This involves iterating through the second half, changing the 'next' pointers to point to the previous node.
Step 4: Initialize 'left' to point to the head of the original linked list and 'right' to point to the head of the reversed second half.
Step 5: Iterate through the first and reversed second halves simultaneously. In each iteration, calculate the sum of 'left.val' and 'right.val', and update 'max_sum' if the current sum is greater.
Step 6: Move 'left' one step forward and 'right' one step forward.
Step 7: Repeat steps 5 and 6 until 'right' reaches the end (null).
Step 8: Return the 'max_sum'.

Key Insights

Insight 1: The problem can be solved by finding the middle of the linked list and reversing the second half.
Insight 2: Using the two-pointer technique (slow and fast pointers) is efficient for finding the middle node of the linked list.
Insight 3: Reversing the second half of the linked list allows for easy calculation of twin sums by iterating from both ends.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Linked List, Two Pointers, Stack.

Companies

Asked at: josh technology.

Maximum Twin Sum of a Linked List | LeetCode 2130

Medium

Linked List Two Pointers Stack

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 int pairSum(ListNode head) {
        ListNode slow = head, fast = head;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }

        ListNode prev = null;
        ListNode curr = slow;
        while (curr != null) {
            ListNode next_temp = curr.next;
            curr.next = prev;
            prev = curr;
            curr = next_temp;
        }

        int max_sum = 0;
        ListNode left = head, right = prev;
        while (right != null) {
            int current_sum = left.val + right.val;
            if (current_sum > max_sum) {
                max_sum = current_sum;
            }
            left = left.next;
            right = right.next;
        }
        
        return max_sum;
    }
}

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Problem Info

Description

Problem 2130: Maximum Twin Sum of a Linked List - Detailed explanation and solution approaches.

Topics

Linked List Two Pointers Stack

Companies

josh technology

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2130

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

8 steps

1
Step 1: Initialize two pointers, 'slow' and 'fast', both pointing to the head of the linked list.
2
Step 2: Move 'slow' one step at a time and 'fast' two steps at a time until 'fast' reaches the end of the list (or 'fast.next' is null). At this point, 'slow' will be pointing to the middle of the list.
3
Step 3: Reverse the second half of the linked list starting from 'slow'. This involves iterating through the second half, changing the 'next' pointers to point to the previous node.
4
Step 4: Initialize 'left' to point to the head of the original linked list and 'right' to point to the head of the reversed second half.
5
Step 5: Iterate through the first and reversed second halves simultaneously. In each iteration, calculate the sum of 'left.val' and 'right.val', and update 'max_sum' if the current sum is greater.
6
Step 6: Move 'left' one step forward and 'right' one step forward.
7
Step 7: Repeat steps 5 and 6 until 'right' reaches the end (null).
8
Step 8: Return the 'max_sum'.

💡

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 traverse the linked list a constant number of times. The space complexity is constant because we only use a few pointer variables that don'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

Using a Stack

Traverse the linked list and push the first half's values onto a stack. Then traverse the second half, popping values from the stack and calculating the twin sum with the current node's value. Keep track of the maximum sum.

Time: O(n)Space: O(n/2) = O(n)

Trade-offs: Using a stack has the same time complexity O(n), but it requires O(n) space because the stack will store the values of the first half of the linked list. The current solution uses O(1) space but modifies the linked list. If modifying the linked list is not allowed, the stack approach would be preferable despite its increased space usage.

✨

Key Insights

1Insight 1: The problem can be solved by finding the middle of the linked list and reversing the second half.
2Insight 2: Using the two-pointer technique (slow and fast pointers) is efficient for finding the middle node of the linked list.
3Insight 3: Reversing the second half of the linked list allows for easy calculation of twin sums by iterating from both ends.

📚

Key Terms Explained

For Students

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

Linked List

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

Edge Cases

Edge case 1: Linked list with only two nodes. The algorithm correctly finds the middle (which is the second node), reverses the second half (which is just swapping the next pointer of the second node to null), and then calculates the twin sum.
Edge case 2: Very large linked list. The algorithm's O(n) time complexity ensures it will process large lists without exceeding linear time. The O(1) space complexity guarantees it won't run out of memory, even for huge inputs.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where the fast pointer reaches the end of the list while finding the middle. Make sure to check for both 'fast' and 'fast.next' being null in the while loop condition.

✗Common mistake 2: Forgetting to reset the 'next' pointer of the last node in the reversed second half to null, potentially creating a circular linked list or causing issues when iterating through the list later.

✅

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 where this algorithm/pattern is useful: Checking if a linked list is a palindrome. You find the middle, reverse the second half, and then compare the first half with the reversed second half.

🏦Real-world scenario 2 with practical application: Memory management in systems where you might need to quickly reverse a portion of a linked list representing available memory blocks.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

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

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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: October 29, 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

❓

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.

When is the two-pointer technique most effective?

Two pointers work best on sorted arrays or when searching for pairs/triplets that satisfy a condition. Common patterns include: moving pointers inward from both ends (like in two-sum variants), fast/slow pointers for cycle detection, or left/right pointers for partitioning.

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 "Maximum Twin Sum of a Linked 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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