What is the time complexity of Longest Palindromic Substring?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Longest Palindromic Substring?

This problem uses Two Pointers, String, Dynamic Programming. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Longest Palindromic Substring?

Insight 1: Palindromes can be centered around a single character (odd length) or between two characters (even length). Insight 2: Expanding around the center is an efficient way to find palindromes.

Which companies ask LeetCode 5. Longest Palindromic Substring in interviews?

This problem is frequently asked at Accenture, Accolite, Adobe, Agoda, Amazon and 68 more companies.

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Longest Palindromic Substring - LeetCode 5 Solution

Q: How to solve Longest Palindromic Substring?

The solution employs the 'expand around center' approach. It iterates through each character in the string and considers it as a potential center of a palindrome. For each character, it expands outwards, checking for both odd-length and even-length palindromes. It maintains the start index and maximum length of the longest palindrome found so far. The core idea is that every palindrome has a center (either a single character or the space between two characters), so we can efficiently find palindromes by expanding from each potential center.

Longest Palindromic Substring - Complete Solution Guide

Longest Palindromic Substring is LeetCode problem 5, 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 a string s , return the longest palindromic substring in s . Example 1: Input: s = "babad" Output: "bab" Explanation: "aba" is also a valid answer. Example 2: Input: s = "cbbd" Output: "bb" Constraints: 1 <= s.length <= 1000 s consist of only digits and English letters.

Detailed Explanation

The problem asks us to find the longest palindromic substring within a given string 's'. A palindrome is a string that reads the same forwards and backward (e.g., "madam", "racecar"). A substring is a contiguous sequence of characters within a string. We need to return this longest palindromic substring. The input string 's' consists of digits and English letters, and its length is between 1 and 1000.

Solution Approach

The solution employs the 'expand around center' approach. It iterates through each character in the string and considers it as a potential center of a palindrome. For each character, it expands outwards, checking for both odd-length and even-length palindromes. It maintains the start index and maximum length of the longest palindrome found so far. The core idea is that every palindrome has a center (either a single character or the space between two characters), so we can efficiently find palindromes by expanding from each potential center.

Step-by-Step Algorithm

Step 1: Initialize variables 'start' to 0 (start index of the longest palindrome), and 'max_len' to 1 (minimum length of a palindrome, i.e., a single character).
Step 2: Iterate through the input string 's' from left to right (index 'i').
Step 3: For each index 'i', consider it as the center of a potential odd-length palindrome. Initialize 'l' and 'r' to 'i'. Expand outwards while 'l' >= 0 and 'r' < n (string length) and s[l] == s[r]. If a longer palindrome is found, update 'start' and 'max_len'.
Step 4: For each index 'i', consider it and the next character 'i+1' as the center of a potential even-length palindrome. Initialize 'l' to 'i' and 'r' to 'i+1'. Expand outwards while 'l' >= 0 and 'r' < n and s[l] == s[r]. If a longer palindrome is found, update 'start' and 'max_len'.
Step 5: After iterating through the entire string, extract the longest palindromic substring using 'start' and 'max_len' (s[start:start + max_len]).

Key Insights

Insight 1: Palindromes can be centered around a single character (odd length) or between two characters (even length).
Insight 2: Expanding around the center is an efficient way to find palindromes.
Insight 3: The optimal approach avoids quadratic space complexity by not storing palindrome lengths in a separate table.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(1)

Topics

This problem involves: Two Pointers, String, Dynamic Programming.

Companies

Asked at: Accenture, Accolite, Adobe, Agoda, Amazon, Apple, Autodesk, BNY Mellon, BlackRock, Bloomberg, ByteDance, CEDCOSS, Cisco, Cognizant, Commvault, Dell, Deloitte, Disney, DoorDash, EPAM Systems, EarnIn, Goldman Sachs, Grab, Huawei, IBM, Info Edge, Infosys, J.P. Morgan, LinkedIn, LiveRamp, MAQ Software, MakeMyTrip, Mastercard, Media.net, Meta, Microsoft, Morgan Stanley, Nielsen, Nutanix, Oracle, Palo Alto Networks, Paytm, PhonePe, Pure Storage, Pwc, RBC, SAP, Salesforce, Samsung, ServiceNow, Softwire, Tesla, ThoughtWorks, TikTok, Tinkoff, Turing, Turo, Uber, UiPath, VMware, Visa, Walmart Labs, Wayfair, Wix, Yahoo, Yandex, Zepto, Zoho, Zopsmart, athenahealth, eBay, persistent systems, tcs.

Longest Palindromic Substring | LeetCode 5

Medium

Two Pointers String Dynamic Programming

Time: O(n^2)Space: O(1)

Solution Code

Available in:

class Solution {
    public String longestPalindrome(String s) {
        int n = s.length();
        if (n < 2) {
            return s;
        }

        int start = 0;
        int max_len = 1;

        for (int i = 0; i < n; i++) {
            // Case 1: Odd length palindrome (center is s[i])
            int l = i, r = i;
            while (l >= 0 && r < n && s.charAt(l) == s.charAt(r)) {
                if ((r - l + 1) > max_len) {
                    start = l;
                    max_len = r - l + 1;
                }
                l -= 1;
                r += 1;
            }

            // Case 2: Even length palindrome (center is between s[i] and s[i+1])
            l = i;
            r = i + 1;
            while (l >= 0 && r < n && s.charAt(l) == s.charAt(r)) {
                if ((r - l + 1) > max_len) {
                    start = l;
                    max_len = r - l + 1;
                }
                l -= 1;
                r += 1;
            }
        }

        return s.substring(start, start + max_len);
    }
}

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Problem Info

Description

Problem 5: Longest Palindromic Substring - Detailed explanation and solution approaches.

Topics

Two Pointers String Dynamic Programming

Companies

AccentureAccoliteAdobeAgodaAmazonAppleAutodeskBNY Mellon

Stats

Difficulty:Medium

Languages:4

Companies:73

Problem #:5

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 variables 'start' to 0 (start index of the longest palindrome), and 'max_len' to 1 (minimum length of a palindrome, i.e., a single character).
2
Step 2: Iterate through the input string 's' from left to right (index 'i').
3
Step 3: For each index 'i', consider it as the center of a potential odd-length palindrome. Initialize 'l' and 'r' to 'i'. Expand outwards while 'l' >= 0 and 'r' < n (string length) and s[l] == s[r]. If a longer palindrome is found, update 'start' and 'max_len'.
4
Step 4: For each index 'i', consider it and the next character 'i+1' as the center of a potential even-length palindrome. Initialize 'l' to 'i' and 'r' to 'i+1'. Expand outwards while 'l' >= 0 and 'r' < n and s[l] == s[r]. If a longer palindrome is found, update 'start' and 'max_len'.
5
Step 5: After iterating through the entire string, extract the longest palindromic substring using 'start' and 'max_len' (s[start:start + max_len]).

💡

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^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n^2) because, in the worst case, the 'expand around center' approach may iterate through a large portion of the string for each character as a potential center. The outer loop iterates 'n' times. The inner 'while' loops, expanding around the center, can iterate up to 'n/2' times in the worst case. Thus, the total time complexity is O(n * n/2) which simplifies to O(n^2). The space complexity is O(1) because we only use a few constant extra variables ('start', 'max_len', 'l', 'r') regardless of the input string's size. The C version needs to allocate memory for the string, which is O(n), but it is still not dependent on any table or data structure related to the length of the string.

📖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

Dynamic Programming

Create a 2D boolean array dp[i][j] that is true if the substring s[i...j] is a palindrome, and false otherwise. Build the table bottom-up, using the relation: dp[i][j] = (s[i] == s[j] && (j-i <= 2 || dp[i+1][j-1])).

Time: O(n^2)Space: O(n^2)

Trade-offs: DP uses O(n^2) space, which is a disadvantage compared to the O(1) space of the 'expand around center' approach. However, DP can be more intuitive for some people to understand and implement. DP might be preferred when the input string 's' may be mutated, and we want to keep track of all palindrome substrings for later retrieval.

Manacher's Algorithm

Manacher's algorithm is an optimized approach that finds the longest palindromic substring in linear time. It preprocesses the string by inserting special characters between each character to handle even and odd length palindromes uniformly.

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

Trade-offs: Manacher's Algorithm achieves O(n) time complexity at the cost of increased complexity in implementation. It's generally harder to understand and implement than the 'expand around center' method or dynamic programming. It's ideal when extremely high performance is required and worth the additional coding effort.

✨

Key Insights

1Insight 1: Palindromes can be centered around a single character (odd length) or between two characters (even length).
2Insight 2: Expanding around the center is an efficient way to find palindromes.
3Insight 3: The optimal approach avoids quadratic space complexity by not storing palindrome lengths in a separate table.

📚

Key Terms Explained

For Students

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty string (n=0). The code handles this by returning the string itself, which is an empty string.
Edge case 2: Single-character string (n=1). The code handles this by returning the string itself, which is the single character.
Edge case 3: String with no palindromic substring longer than 1. The algorithm correctly identifies a single character as the longest palindrome.
Edge case 4: String containing only the same character (e.g., "aaaa"). The algorithm correctly identifies the entire string as the longest palindrome.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to consider both odd and even length palindromes.

✗Common mistake 2: Incorrectly updating 'start' and 'max_len' when a longer palindrome is found. Ensure 'start' is updated to the leftmost index of the new longest palindrome.

✗Common mistake 3: Off-by-one errors when extracting the substring. Pay attention to the start index and the length of the substring.

✅

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: Bioinformatics: Identifying palindromic sequences in DNA or RNA, which are often associated with regulatory regions or structural elements.

🏦Real-world scenario 2: Text editing/processing: Implementing features like autocorrect or spellcheck, where identifying palindromic errors can be useful.

🎮Real-world scenario 3: Data compression: Palindromic patterns can be exploited for data compression techniques.

📱Real-world scenario 4: Computer Vision: Recognizing symmetric patterns in images, such as faces or objects.

💡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 two-pointer technique is powerful for sorted arrays. The key insight is recognizing when elements can be processed from both ends simultaneously.

🎯 Interview Tip

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

⚠️ Watch Out

The most common mistake is not handling the case when pointers meet. Always consider the termination condition.

👨‍💻 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: June 6, 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 do I avoid infinite loops with two pointers?

Always ensure your loop condition guarantees termination (typically left < right or slow != fast). Move at least one pointer in each iteration, and handle edge cases where pointers might not move (like skipping duplicates). Draw out pointer movement on paper first.

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 "Longest Palindromic Substring"?

For this medium-level Two Pointers problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?