What is the time complexity of Palindromic Substrings?

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

What data structures are used to solve Palindromic Substrings?

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

What are the key insights for solving Palindromic Substrings?

Insight 1: Every single character in the string is a palindrome of length 1. Insight 2: Palindromes can be centered around a single character or between two characters. For example, "aba" is centered around 'b', while "aa" is centered between the two 'a's.

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Palindromic Substrings - LeetCode 647 Solution

Q: How to solve Palindromic Substrings?

The solution uses the 'Expand Around Center' approach. This method iterates through all possible centers of palindromes. Since a palindrome can be centered around a single character or between two characters, we have `2 * n - 1` possible centers for a string of length `n`. For each center, we expand outwards, checking if the characters to the left and right are equal. If they are, we increment the count of palindromic substrings and continue expanding. If they are not, we stop expanding from that center. This approach efficiently identifies all palindromes without requiring a separate palindrome check function.

Q: Which companies ask LeetCode 647. Palindromic Substrings in interviews?

This problem is frequently asked at Accenture, Arista Networks, BNY Mellon, Cisco, Citadel and 9 more companies.

Palindromic Substrings - Complete Solution Guide

Palindromic Substrings is LeetCode problem 647, 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 number of palindromic substrings in it. A string is a palindrome when it reads the same backward as forward. A substring is a contiguous sequence of characters within the string. Example 1: Input: s = "abc" Output: 3 Explanation: Three palindromic strings: "a", "b", "c". Example 2: Input: s = "aaa" Output: 6 Explanation: Six palindromic strings: "a", "a", "a", "aa", "aa", "aaa". Constraints: 1 <= s.length <= 1000 s consists of lowercase English letters.

Detailed Explanation

The problem asks us to find the total number of palindromic substrings within a given string `s`. A palindrome is a sequence that reads the same forwards and backward. A substring is a contiguous sequence of characters within the string. For example, in the string "abc", the palindromic substrings are "a", "b", and "c". In the string "aaa", the palindromic substrings are "a", "a", "a", "aa", "aa", and "aaa". The length of the string `s` is constrained to be between 1 and 1000, and it contains only lowercase English letters.

Solution Approach

The solution uses the 'Expand Around Center' approach. This method iterates through all possible centers of palindromes. Since a palindrome can be centered around a single character or between two characters, we have `2 * n - 1` possible centers for a string of length `n`. For each center, we expand outwards, checking if the characters to the left and right are equal. If they are, we increment the count of palindromic substrings and continue expanding. If they are not, we stop expanding from that center. This approach efficiently identifies all palindromes without requiring a separate palindrome check function.

Step-by-Step Algorithm

Step 1: Initialize a variable `res` (or `count`) to 0, which will store the total number of palindromic substrings.
Step 2: Iterate through all possible centers. The loop goes from `center = 0` to `2 * n - 2` where n is the length of the string.
Step 3: For each center, determine the left and right pointers. `left = center // 2` and `right = left + (center % 2)`. If `center` is even, the palindrome is centered at a single character, and if it's odd, it's centered between two characters.
Step 4: Expand outwards from the center using a `while` loop. The loop continues as long as `left` and `right` are within the bounds of the string (`left >= 0` and `right < n`) and the characters at the left and right pointers are equal (`s[left] == s[right]`).
Step 5: Inside the `while` loop, increment `res` (or `count`) because we have found a new palindromic substring. Decrement `left` and increment `right` to expand outwards.
Step 6: After the `while` loop finishes, move to the next center and repeat steps 3-5.
Step 7: Finally, return `res` (or `count`), which represents the total number of palindromic substrings in the string.

Key Insights

Insight 1: Every single character in the string is a palindrome of length 1.
Insight 2: Palindromes can be centered around a single character or between two characters. For example, "aba" is centered around 'b', while "aa" is centered between the two 'a's.
Insight 3: An efficient approach is to expand outwards from each possible center and check if the expanding substring is a palindrome.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(1)

Topics

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

Companies

Asked at: Accenture, Arista Networks, BNY Mellon, Cisco, Citadel, Epic Systems, LinkedIn, Millennium, Netskope, PayPal, Pure Storage, Salesforce, SoFi, Wayfair.

Palindromic Substrings | LeetCode 647

Medium

String Dynamic Programming Two Pointers

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

Solution Code

Available in:

class Solution {
    public int countSubstrings(String s) {
        int n = s.length();
        int count = 0;
        for (int center = 0; center < 2 * n - 1; center++) {
            int left = center / 2;
            int right = left + (center % 2);
            while (left >= 0 && right < n && s.charAt(left) == s.charAt(right)) {
                count++;
                left--;
                right++;
            }
        }
        return count;
    }
}

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Problem Info

Description

Problem 647: Palindromic Substrings - Detailed explanation and solution approaches.

Topics

String Dynamic Programming Two Pointers

Companies

AccentureArista NetworksBNY MellonCiscoCitadelEpic SystemsLinkedInMillennium

Stats

Difficulty:Medium

Languages:4

Companies:14

Problem #:647

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: Initialize a variable `res` (or `count`) to 0, which will store the total number of palindromic substrings.
2
Step 2: Iterate through all possible centers. The loop goes from `center = 0` to `2 * n - 2` where n is the length of the string.
3
Step 3: For each center, determine the left and right pointers. `left = center // 2` and `right = left + (center % 2)`. If `center` is even, the palindrome is centered at a single character, and if it's odd, it's centered between two characters.
4
Step 4: Expand outwards from the center using a `while` loop. The loop continues as long as `left` and `right` are within the bounds of the string (`left >= 0` and `right < n`) and the characters at the left and right pointers are equal (`s[left] == s[right]`).
5
Step 5: Inside the `while` loop, increment `res` (or `count`) because we have found a new palindromic substring. Decrement `left` and increment `right` to expand outwards.
6
Step 6: After the `while` loop finishes, move to the next center and repeat steps 3-5.
7
Step 7: Finally, return `res` (or `count`), which represents the total number of palindromic substrings in the string.

💡

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 outer loop iterates `2 * n - 1` times (O(n)). The inner `while` loop, in the worst case, can iterate up to `n/2` times. Thus the overall time complexity is O(n * n/2), which simplifies to O(n^2). The algorithm uses only a constant amount of extra space for variables like `left`, `right`, and `res`, so the space complexity is O(1).

📖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

A dynamic programming approach can also be used to solve this problem. We can create a 2D boolean array `dp[i][j]` where `dp[i][j]` is true if the substring `s[i...j]` is a palindrome, and false otherwise. The recurrence relation is 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: The dynamic programming approach has the same time complexity as the expand-around-center approach but uses O(n^2) space. The expand-around-center approach is usually preferred because of its lower space complexity. DP might be preferred if you need to repeatedly query whether a given substring is a palindrome.

✨

Key Insights

1Insight 1: Every single character in the string is a palindrome of length 1.
2Insight 2: Palindromes can be centered around a single character or between two characters. For example, "aba" is centered around 'b', while "aa" is centered between the two 'a's.
3Insight 3: An efficient approach is to expand outwards from each possible center and check if the expanding substring is a palindrome.

📚

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. The code correctly handles an empty string because the loop won't execute, and 0 will be returned.
Edge case 2: Single-character string. The code will correctly identify the single character as a palindrome, returning 1.
Edge case 3: String with all same characters, e.g., "aaa". The expansion will correctly identify all possible palindromic substrings including "a", "a", "a", "aa", "aa", and "aaa".

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering both single-character and between-character centers for palindromes. This would result in missing palindromes like 'aa' or 'abba'.

✗Common mistake 2: Incorrectly defining the loop conditions for expanding outwards. Ensure that the left and right pointers stay within the bounds of the string.

✗Common mistake 3: Forgetting to increment the count of palindromes within the while loop while expanding outwards.

✅

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 - Finding palindromic sequences in DNA, which can indicate regulatory elements.

🏦Real-world scenario 2: Text processing - Palindrome detection can be useful in text analysis tasks like identifying interesting patterns or symmetric structures in language. It could also be used in data compression algorithms.

💡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

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

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

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

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 16, 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

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

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 30

❓

Frequently Asked Questions

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.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

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 "Palindromic Substrings"?

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

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