What is the time complexity of Palindrome Linked List?

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

What data structures are used to solve Palindrome Linked List?

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

What are the key insights for solving Palindrome Linked List?

Insight 1: The most efficient way to solve this problem is by using two pointers: a slow pointer and a fast pointer. The fast pointer moves twice as fast as the slow pointer. When the fast pointer reaches the end, the slow pointer will be at the middle of the list. Insight 2: To check if the first half and second half are equal, reverse the second half of the linked list. Then compare the first half and the reversed second half element by element.

Which companies ask LeetCode 234. Palindrome Linked List in interviews?

This problem is frequently asked at Barclays, Devsinc, IXL, Intuit, Morgan Stanley and 6 more companies.

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

Palindrome Linked List - LeetCode 234 Solution

Palindrome Linked List - Complete Solution Guide

Palindrome Linked List is LeetCode problem 234, a Easy 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, return true if it is a palindrome or false otherwise . Example 1: Input: head = [1,2,2,1] Output: true Example 2: Input: head = [1,2] Output: false Constraints: The number of nodes in the list is in the range [1, 10 5 ] . 0 <= Node.val <= 9 Follow up: Could you do it in O(n) time and O(1) space?

Detailed Explanation

The problem asks to determine if a given singly linked list is a palindrome. A palindrome is a sequence that reads the same backward as forward. The input is the head of the linked list, and the output is a boolean value: `true` if the linked list is a palindrome, and `false` otherwise. The list nodes contain integer values between 0 and 9 inclusive. The list can have between 1 and 10<sup>5</sup> nodes.

Solution Approach

The solution employs a two-pointer approach to find the middle of the linked list. Then, it reverses the second half of the list and compares it with the first half. If the values match, the linked list is a palindrome; otherwise, it is not.

Step-by-Step Algorithm

Step 1: Initialize two pointers, `slow` and `fast`, both pointing to the head of the linked list.
Step 2: Move `fast` two steps forward and `slow` one step forward in each iteration until `fast` reaches the end of the list. At this point, `slow` will be pointing to the middle node of the list.
Step 3: Reverse the second half of the list starting from the node pointed by `slow` using iterative reversal.
Step 4: Compare the values of the nodes in the first half and the reversed second half. If any pair of values doesn't match, return `false`. Otherwise, continue comparison.
Step 5: If all the node values match, return `true` indicating the list is a palindrome.

Key Insights

Insight 1: The most efficient way to solve this problem is by using two pointers: a slow pointer and a fast pointer. The fast pointer moves twice as fast as the slow pointer. When the fast pointer reaches the end, the slow pointer will be at the middle of the list.
Insight 2: To check if the first half and second half are equal, reverse the second half of the linked list. Then compare the first half and the reversed second half element by element.
Insight 3: The solution must handle edge cases like empty lists, lists with one node, and lists with even or odd numbers of nodes. The algorithm cleverly handles all these by using the fast pointer condition `fast && fast->next` (or equivalent in other languages).

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

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

Companies

Asked at: Barclays, Devsinc, IXL, Intuit, Morgan Stanley, Oracle, ServiceNow, Siemens, Yandex, ZScaler, Zoho.

Palindrome Linked List | LeetCode 234

Easy

Linked List Two Pointers Stack 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 boolean isPalindrome(ListNode head) {
        ListNode slow = head;
        ListNode fast = head;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        ListNode secondHalf = reverseList(slow);
        while (secondHalf != null) {
            if (head.val != secondHalf.val) {
                return false;
            }
            head = head.next;
            secondHalf = secondHalf.next;
        }
        return true;
    }
    private ListNode reverseList(ListNode head) {
        ListNode prev = null;
        while (head != null) {
            ListNode next = head.next;
            head.next = prev;
            prev = head;
            head = next;
        }
        return prev;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 234: Palindrome Linked List - Detailed explanation and solution approaches.

Topics

Linked List Two Pointers Stack Recursion

Companies

BarclaysDevsincIXLIntuitMorgan StanleyOracleServiceNowSiemens

Stats

Difficulty:Easy

Languages:4

Companies:11

Problem #:234

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

5 steps

1
Step 1: Initialize two pointers, `slow` and `fast`, both pointing to the head of the linked list.
2
Step 2: Move `fast` two steps forward and `slow` one step forward in each iteration until `fast` reaches the end of the list. At this point, `slow` will be pointing to the middle node of the list.
3
Step 3: Reverse the second half of the list starting from the node pointed by `slow` using iterative reversal.
4
Step 4: Compare the values of the nodes in the first half and the reversed second half. If any pair of values doesn't match, return `false`. Otherwise, continue comparison.
5
Step 5: If all the node values match, return `true` indicating the list is a palindrome.

💡

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 algorithm traverses the list once to find the middle (O(n)), reverses the second half (O(n)), and compares the two halves (O(n)). The dominant factor is O(n). Space complexity is O(1) because only a few constant extra variables are used; no additional data structures that scale with the input size are employed.

📖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 once and push each node's value onto a stack. Then, traverse the list again and compare each node's value with the top of the stack. Pop the stack after each comparison.

Time: O(n)Space: O(n)

Trade-offs: This approach is simpler to understand, but it uses O(n) extra space to store the stack, unlike the optimal solution's O(1) space complexity.

Recursion

Recursively check if the list is a palindrome by comparing the first and last nodes and then recursively calling the function for the rest of the list.

Time: O(n)Space: O(n)

Trade-offs: Recursive approach can be more readable but uses O(n) space due to recursive call stack. Not as efficient as the iterative two-pointer approach.

✨

Key Insights

1Insight 1: The most efficient way to solve this problem is by using two pointers: a slow pointer and a fast pointer. The fast pointer moves twice as fast as the slow pointer. When the fast pointer reaches the end, the slow pointer will be at the middle of the list.
2Insight 2: To check if the first half and second half are equal, reverse the second half of the linked list. Then compare the first half and the reversed second half element by element.
3Insight 3: The solution must handle edge cases like empty lists, lists with one node, and lists with even or odd numbers of nodes. The algorithm cleverly handles all these by using the fast pointer condition `fast && fast->next` (or equivalent in other languages).

📚

Key Terms Explained

For Students

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Recursion

A function that calls itself to solve smaller versions of the same problem.

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: Empty list or list with one node. The solution correctly handles these cases as the while loop conditions (`fast && fast->next` or equivalent) will not execute, and the subsequent comparison will trivially return `true` (for a single-node list) or no comparison happens (empty list) and `true` is implied.
Edge case 2: List with even number of nodes. The slow pointer will stop at the middle of the second half of the list, which is correctly handled by the algorithm.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling edge cases correctly, especially the empty list or list with a single node. Incorrect loop conditions can lead to errors here.

✗Common mistake 2: Incorrectly reversing the second half of the linked list. A common error is mixing up the order of pointer assignments in the reversal process, leading to a corrupted 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: Checking for palindromic sequences in biological data (e.g., DNA sequences) can leverage a linked list representation and this algorithm.

🏦Real-world scenario 2: Network routing protocols might use linked lists to represent paths. Palindrome checks (though less direct) could be used in debugging or path optimization to detect symmetrical or redundant connections.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ 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: July 25, 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.

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.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

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 "Palindrome Linked List"?

For this easy-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.

🤖 Need help with this problem?