What is the time complexity of Lexicographically Smallest Palindrome?

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

What data structures are used to solve Lexicographically Smallest Palindrome?

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

What are the key insights for solving Lexicographically Smallest Palindrome?

Insight 1: The solution only needs to focus on the first half of the string. The second half is automatically determined to create a palindrome. Insight 2: A two-pointer approach (one at the beginning and one at the end) is efficient for comparing and modifying characters to create a palindrome.

Which companies ask LeetCode 2697. Lexicographically Smallest Palindrome in interviews?

This problem is frequently asked at PayPal.

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Lexicographically Smallest Palindrome - LeetCode 2697 Solution

Lexicographically Smallest Palindrome - Complete Solution Guide

Lexicographically Smallest Palindrome is LeetCode problem 2697, 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

You are given a string s consisting of lowercase English letters , and you are allowed to perform operations on it. In one operation, you can replace a character in s with another lowercase English letter. Your task is to make s a palindrome with the minimum number of operations possible. If there are multiple palindromes that can be made using the minimum number of operations, make the lexicographically smallest one. A string a is lexicographically smaller than a string b (of the same length) i

Detailed Explanation

The problem asks you to transform a given string `s` consisting of lowercase English letters into its lexicographically smallest palindrome using the minimum number of character replacements. A palindrome is a string that reads the same forwards and backward. The lexicographically smallest palindrome means that among all possible palindromes achievable with the minimum number of changes, you should choose the one that comes earliest in alphabetical order. For example, "abba" is lexicographically smaller than "bbaa". The input is a string `s`, and the output is the lexicographically smallest palindrome that can be created from `s`.

Solution Approach

The provided code utilizes a two-pointer approach. It iterates through the input string, comparing characters from the beginning and end, moving towards the middle. If a pair of characters is different, the smaller character is chosen to replace the larger character to minimize the number of changes and ensure lexicographical order. This guarantees that we get the smallest lexicographical palindrome with minimal changes.

Step-by-Step Algorithm

Step 1: Convert the input string to a character array for easier manipulation.
Step 2: Initialize two pointers, `left` pointing to the beginning of the array and `right` pointing to the end.
Step 3: Iterate while `left` is less than `right`. In each iteration:
Step 4: Compare `chars[left]` and `chars[right]`. If they are different, replace the larger character with the smaller character.
Step 5: Increment `left` and decrement `right`, moving the pointers closer to the middle.
Step 6: Once the loop finishes, convert the modified character array back to a string and return it.

Key Insights

Insight 1: The solution only needs to focus on the first half of the string. The second half is automatically determined to create a palindrome.
Insight 2: A two-pointer approach (one at the beginning and one at the end) is efficient for comparing and modifying characters to create a palindrome.
Insight 3: To create the lexicographically smallest palindrome, always choose the smaller character when a pair of characters differs, replacing the larger character with the smaller one.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Two Pointers, String, Greedy.

Companies

Asked at: PayPal.

Lexicographically Smallest Palindrome | LeetCode 2697

Easy

Two Pointers String Greedy

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

Solution Code

Available in:

class Solution {
    public String makeSmallestPalindrome(String s) {
        char[] chars = s.toCharArray();
        int left = 0;
        int right = chars.length - 1;

        while (left < right) {
            if (chars[left] != chars[right]) {
                if (chars[left] < chars[right]) {
                    chars[right] = chars[left];
                } else {
                    chars[left] = chars[right];
                }
            }
            left++;
            right--;
        }

        return new String(chars);
    }
}

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Problem Info

Description

Problem 2697: Lexicographically Smallest Palindrome - Detailed explanation and solution approaches.

Topics

Two Pointers String Greedy

Companies

PayPal

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:2697

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: Convert the input string to a character array for easier manipulation.
2
Step 2: Initialize two pointers, `left` pointing to the beginning of the array and `right` pointing to the end.
3
Step 3: Iterate while `left` is less than `right`. In each iteration:
4
Step 4: Compare `chars[left]` and `chars[right]`. If they are different, replace the larger character with the smaller character.
5
Step 5: Increment `left` and decrement `right`, moving the pointers closer to the middle.
6
Step 6: Once the loop finishes, convert the modified character array back to a string and return it.

💡

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 algorithm iterates through the string once (or half the string in some implementations), resulting in a linear time complexity. The space complexity is constant because in-place modifications are done on the character array; it doesn't scale with the input size. Some versions might use a temporary character array which will slightly increase space complexity, but it will still be 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

✨

Key Insights

1Insight 1: The solution only needs to focus on the first half of the string. The second half is automatically determined to create a palindrome.
2Insight 2: A two-pointer approach (one at the beginning and one at the end) is efficient for comparing and modifying characters to create a palindrome.
3Insight 3: To create the lexicographically smallest palindrome, always choose the smaller character when a pair of characters differs, replacing the larger character with the smaller one.

📚

Key Terms Explained

For Students

Two Pointers

Using two indices to traverse data from different positions simultaneously.

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: Empty string: The code handles empty strings gracefully because the while loop condition (`left < right`) won't be met, and an empty string will be returned.
Edge case 2: Palindrome Input: If the input is already a palindrome, no changes will be made, and the function correctly returns the original string.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case where the input string is already a palindrome or empty. The provided code handles these edge cases correctly.

✗Common mistake 2: Incorrectly choosing the character to replace. Always select the smaller character to ensure the lexicographically smallest palindrome.

✅

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: Data compression. Finding palindromic patterns in data can be helpful for data compression algorithms.

🏦Real-world scenario 2: Bioinformatics. Sequence alignment and analysis in genomics often involves finding palindromic sequences, which are biologically significant.

💡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

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

⚠️ Watch Out

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

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: November 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

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

@InterviewProJul 13

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

@CodeNinjaJul 6

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

@TechLeadJun 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.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 "Lexicographically Smallest Palindrome"?

For this easy-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.

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