What is the time complexity of Break a Palindrome?

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

What data structures are used to solve Break a Palindrome?

This problem uses String, Greedy. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Break a Palindrome?

Insight 1: To achieve the lexicographically smallest non-palindrome, try to replace the leftmost non-'a' character with 'a'. This prioritizes the beginning of the string. Insight 2: Only the first half of the palindrome needs to be considered because any change made in the first half will also affect the second half due to its palindromic nature.

Which companies ask LeetCode 1328. Break a Palindrome in interviews?

This problem is frequently asked at Expedia, J.P. Morgan, MathWorks, Nvidia, VMware.

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

Break a Palindrome - LeetCode 1328 Solution

Break a Palindrome - Complete Solution Guide

Break a Palindrome is LeetCode problem 1328, 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 palindromic string of lowercase English letters palindrome , replace exactly one character with any lowercase English letter so that the resulting string is not a palindrome and that it is the lexicographically smallest one possible. Return the resulting string. If there is no way to replace a character to make it not a palindrome, return an empty string . A string a is lexicographically smaller than a string b (of the same length) if in the first position where a and b differ, a has a c

Detailed Explanation

The problem requires you to modify a given palindromic string by replacing exactly one character with another lowercase English letter such that the resulting string is no longer a palindrome and is the lexicographically smallest possible string after the modification. If no such replacement is possible (e.g., the input is a single character), return an empty string.

Solution Approach

The solution iterates through the first half of the palindrome string. If it encounters a character that is not 'a', it replaces that character with 'a' and returns the modified string. If all characters in the first half are 'a', it means the palindrome consists of 'a's (or mostly 'a's with a single middle character). In this case, the last character of the string is replaced with 'b' to break the palindrome. If the string length is 1, there's no way to break the palindrome, so an empty string is returned.

Step-by-Step Algorithm

Step 1: Check if the length of the input `palindrome` is less than or equal to 1. If it is, return an empty string because a single-character palindrome cannot be broken.
Step 2: Iterate through the first half of the `palindrome` string (up to `n // 2`).
Step 3: Inside the loop, check if the character at the current index `i` is not equal to 'a'.
Step 4: If the character at index `i` is not 'a', replace it with 'a'. Return the modified string immediately. This ensures that the lexicographically smallest non-palindrome is generated.
Step 5: If the loop completes without finding a character to replace with 'a', it means the first half of the palindrome (and thus most of the string) consists only of 'a's. In this case, change the last character of the string to 'b'.
Step 6: Return the modified string.

Key Insights

Insight 1: To achieve the lexicographically smallest non-palindrome, try to replace the leftmost non-'a' character with 'a'. This prioritizes the beginning of the string.
Insight 2: Only the first half of the palindrome needs to be considered because any change made in the first half will also affect the second half due to its palindromic nature.
Insight 3: A special case exists when all characters in the first half of the palindrome are 'a'. In this scenario, changing the last character to 'b' ensures the string is no longer a palindrome.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: String, Greedy.

Companies

Asked at: Expedia, J.P. Morgan, MathWorks, Nvidia, VMware.

Break a Palindrome | LeetCode 1328

Medium

String Greedy

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

Solution Code

Available in:

class Solution {
    public String breakPalindrome(String palindrome) {
        int n = palindrome.length();
        if (n <= 1) {
            return "";
        }
        
        for (int i = 0; i < n / 2; i++) {
            if (palindrome.charAt(i) != 'a') {
                return palindrome.substring(0, i) + 'a' + palindrome.substring(i + 1);
            }
        }
        
        return palindrome.substring(0, n - 1) + 'b';
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 1328: Break a Palindrome - Detailed explanation and solution approaches.

Topics

String Greedy

Companies

ExpediaJ.P. MorganMathWorksNvidiaVMware

Stats

Difficulty:Medium

Languages:4

Companies:5

Problem #:1328

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

6 steps

1
Step 1: Check if the length of the input `palindrome` is less than or equal to 1. If it is, return an empty string because a single-character palindrome cannot be broken.
2
Step 2: Iterate through the first half of the `palindrome` string (up to `n // 2`).
3
Step 3: Inside the loop, check if the character at the current index `i` is not equal to 'a'.
4
Step 4: If the character at index `i` is not 'a', replace it with 'a'. Return the modified string immediately. This ensures that the lexicographically smallest non-palindrome is generated.
5
Step 5: If the loop completes without finding a character to replace with 'a', it means the first half of the palindrome (and thus most of the string) consists only of 'a's. In this case, change the last character of the string to 'b'.
6
Step 6: Return the modified 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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the loop that iterates through the first half of the string. The space complexity depends on whether the string is modified in-place (O(1)) or if a new string needs to be created to store the modified version (O(n)).

📖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

Brute Force

Generate all possible strings by replacing one character at a time and then check if each string is not a palindrome. Find the lexicographically smallest non-palindrome.

Time: O(26*n*n), where n is the length of the string. We iterate through each position (n) and replace it with all 26 possible characters. Then, for each modified string, we check if it is a palindrome (O(n)).Space: O(n), for storing the modified string.

Trade-offs: This approach is much slower than the optimal solution. It should only be considered when the string is very small or as a baseline for comparison. It is less efficient due to generating and checking many unnecessary strings.

✨

Key Insights

1Insight 1: To achieve the lexicographically smallest non-palindrome, try to replace the leftmost non-'a' character with 'a'. This prioritizes the beginning of the string.
2Insight 2: Only the first half of the palindrome needs to be considered because any change made in the first half will also affect the second half due to its palindromic nature.
3Insight 3: A special case exists when all characters in the first half of the palindrome are 'a'. In this scenario, changing the last character to 'b' ensures the string is no longer a palindrome.

📚

Key Terms Explained

For Students

Greedy Algorithm

Making the best choice at each step without looking ahead.

String

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

Edge Cases

Edge case 1: Input is a single character string, like "a". The code correctly returns an empty string.
Edge case 2: Input is a string of all 'a's, like "aaaa". The code correctly changes the last character to 'b' resulting in "aaab".
Edge case 3: Input is a string like "aba". The code changes the first 'a' to 'a', skips because it is equal to 'a' and proceeds to change last character to 'b', resulting in "abb".
Edge case 4: Input is "abba", the code changes it to "aaba".

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the edge case where the input string has length 1 or less, leading to incorrect output or errors.

✗Common mistake 2: Not considering the case where all characters in the first half of the palindrome are 'a'. If this occurs, the solution needs to modify the last character to 'b' to ensure the string is no longer a palindrome.

✗Common mistake 3: Not focusing on the lexicographically smallest solution. Naively replacing any character may result in a non-palindrome but not the smallest one.

✗Common mistake 4: Trying to modify characters in the second half of the palindrome. The modifications can be done only in the first half of the string and ensure the string isn't a palindrome after mirroring the changes to the second half.

✅

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: This algorithm can be used as a building block in systems that generate unique identifiers or passwords based on specific rules and constraints.

🏦Real-world scenario 2: It can be applied in text processing applications where you need to slightly modify strings while maintaining a degree of similarity and minimizing changes, such as in data anonymization or code obfuscation.

💡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: August 8, 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

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

@CodeNinjaJul 14

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

@TechLeadJul 7

❓

Frequently Asked Questions

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How do I verify if a greedy approach is correct?

Prove the greedy choice property: the locally optimal choice leads to globally optimal solution. Use exchange argument: show that swapping any element with the greedy choice doesn't improve the solution. If you can't prove it, test extensively or consider DP.

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 "Break a Palindrome"?

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 greedy techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?