What is the time complexity of Smallest Palindromic Rearrangement I?

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

What data structures are used to solve Smallest Palindromic Rearrangement I?

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

What are the key insights for solving Smallest Palindromic Rearrangement I?

Insight 1: To form the lexicographically smallest palindrome, we should place the smallest characters in the first half of the string and their duplicates in the second half in reverse order. Insight 2: If a character appears an odd number of times, it must be placed in the middle of the palindrome.

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Smallest Palindromic Rearrangement I - LeetCode 3517 Solution

Smallest Palindromic Rearrangement I - Complete Solution Guide

Smallest Palindromic Rearrangement I is LeetCode problem 3517, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Framing

Smallest Palindromic Rearrangement I is a Medium LeetCode problem that rewards careful tracing, edge-case handling, and a clear grasp of String and Sorting. The best solutions usually explain why the chosen invariant holds before they optimize for time or space.

Quick Example Mindset

A useful way to test Smallest Palindromic Rearrangement I is to start with a tiny input that exposes the boundary conditions, then run the same logic on a slightly larger case to verify the string behavior and the sorting interaction. That second pass is where off-by-one mistakes and missing updates usually appear.

Problem Statement

You are given a palindromic string s . Return the lexicographically smallest palindromic permutation of s . Example 1: Input: s = "z" Output: "z" Explanation: A string of only one character is already the lexicographically smallest palindrome. Example 2: Input: s = "babab" Output: "abbba" Explanation: Rearranging "babab" → "abbba" gives the smallest lexicographic palindrome. Example 3: Input: s = "daccad" Output: "acddca" Explanation: Rearranging "daccad" → "acddca" gives the smallest

Detailed Explanation

The problem requires us to find the lexicographically smallest palindromic permutation of a given palindromic string. A palindromic permutation is a rearrangement of the characters of the string such that the resulting string is a palindrome (reads the same forwards and backward). Lexicographically smallest means we want the palindrome that comes earliest in dictionary order. The input is a string `s` consisting of lowercase English letters that is guaranteed to be palindromic. The output is the lexicographically smallest palindromic permutation of `s`.

Solution Approach

The general approach involves counting the occurrences of each character in the string. Then, we identify the character that appears an odd number of times (if any), which will form the middle of the palindrome. Next, we sort the characters based on their lexicographical order. We take half the count of each character and construct the left half of the palindrome. Finally, the right half of the palindrome is simply the reversed left half, and we concatenate them (left + middle + reversed left) to form the final result.

Step-by-Step Algorithm

Step 1: Count the occurrences of each character in the input string `s` using a dictionary or an array.
Step 2: Identify the character that appears an odd number of times. If multiple characters appear an odd number of times, then there's no palindromic permutation which contradicts the prompt. In such cases, pick the first such character encountered. Since it's guaranteed that s is a palindrome, we will always encounter at most one such character.
Step 3: Construct the left half of the palindrome by iterating through the sorted characters and appending half of their counts to the `left` string.
Step 4: Concatenate the left half, the middle character (if any), and the reversed left half to form the lexicographically smallest palindromic permutation.
Step 5: Return the resulting palindrome.

Key Insights

Insight 1: To form the lexicographically smallest palindrome, we should place the smallest characters in the first half of the string and their duplicates in the second half in reverse order.
Insight 2: If a character appears an odd number of times, it must be placed in the middle of the palindrome.
Insight 3: Because the input string is guaranteed to be a palindrome, we can always construct a palindromic permutation.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: String, Sorting, Counting Sort.

Study Paths

Continue from this problem into the surrounding topic and company clusters to compare how the same pattern appears in other interview settings.

Related topics: String, Sorting, Counting Sort

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.

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.

Smallest Palindromic Rearrangement I | LeetCode 3517

Medium

String Sorting Counting Sort

Study Path

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String Sorting Counting Sort

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

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public String smallestPalindrome(String s) {
        char[] chars = s.toCharArray();
        Arrays.sort(chars);
        return new String(chars);
    }
}

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Problem Info

Description

Problem 3517: Smallest Palindromic Rearrangement I - Detailed explanation and solution approaches.

Topics

String Sorting Counting Sort

Stats

Difficulty:Medium

Languages:4

Problem #:3517

Problem Explanation

🧭

Next Steps

Use this problem as a starting point, then move into adjacent patterns and company sets to reinforce the same idea in different contexts.

Topic hub

Study neighboring problems by pattern and difficulty.

Company hub

See where this problem appears in interview loops.

Pattern guides

Read deeper walkthroughs of the algorithm family.

💡

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: Count the occurrences of each character in the input string `s` using a dictionary or an array.
2
Step 2: Identify the character that appears an odd number of times. If multiple characters appear an odd number of times, then there's no palindromic permutation which contradicts the prompt. In such cases, pick the first such character encountered. Since it's guaranteed that s is a palindrome, we will always encounter at most one such character.
3
Step 3: Construct the left half of the palindrome by iterating through the sorted characters and appending half of their counts to the `left` string.
4
Step 4: Concatenate the left half, the middle character (if any), and the reversed left half to form the lexicographically smallest palindromic permutation.
5
Step 5: Return the resulting 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!

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Complexity Analysis

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Time Complexity

O(n log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is dominated by the sorting step (O(n log n)). The space complexity arises from storing the counts of the character. In the Python and C++ solutions where the characters are stored individually, the space complexity is O(n). The Java solution uses `Arrays.sort()` which, depending on the Java implementation, can use O(log n) to O(n) auxiliary space. Thus, overall it's 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

Counting Sort

Instead of using a general-purpose sorting algorithm, we can use counting sort since the characters are lowercase English letters. This would give us O(n+k) time complexity where k is the size of the character set (26 in this case). Thus it amounts to O(n).

Time: O(n)Space: O(1) or O(k) where k is the alphabet size

Trade-offs: Counting sort is more efficient than general-purpose sorting for strings with a small alphabet size. If the alphabet size is very large or unknown, a general-purpose sorting algorithm might be more appropriate.

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Key Insights

1Insight 1: To form the lexicographically smallest palindrome, we should place the smallest characters in the first half of the string and their duplicates in the second half in reverse order.
2Insight 2: If a character appears an odd number of times, it must be placed in the middle of the palindrome.
3Insight 3: Because the input string is guaranteed to be a palindrome, we can always construct a palindromic permutation.

📚

Key Terms Explained

For Students

String

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

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty string. While the constraints say the string will have length between 1 and 10^5, we can handle the empty string gracefully by returning an empty string or throwing an exception. The provided solutions assume input string length >=1.
Edge case 2: String with a single character (e.g., "z"). The algorithm correctly handles this case as the character is already the lexicographically smallest palindrome.
Edge case 3: String with all the same characters (e.g., "aaaaa"). The algorithm handles this case correctly by constructing the palindrome with the repeating character in the middle and filling the remaining positions with the same character.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case where a character appears an odd number of times. The middle character of the palindrome must be handled separately.

✗Common mistake 2: Not sorting the characters lexicographically before constructing the left half. This will result in a palindrome that is not the lexicographically smallest.

✗Common mistake 3: Incorrectly reversing the left half when constructing the right half of the palindrome. Make sure the reversal is done correctly.

✅

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: Text processing and analysis: Finding the smallest palindromic permutation could be relevant in tasks like anagram generation or identifying symmetrical patterns within text data.

🏦Real-world scenario 2: Bioinformatics: Finding palindromic sequences in DNA or RNA is a common task. This algorithm could be adapted to find the lexicographically smallest such sequence, although biological applications might have different criteria for 'smallest'.

🎮Real-world scenario 3: Data Compression: Palindromic subsequences can be used in certain 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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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 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

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

@InterviewProJul 21

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

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Frequently Asked Questions

What is the sliding window technique for string problems?

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

What is the importance of time and space complexity analysis?

What is the optimal approach for "Smallest Palindromic Rearrangement I"?

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

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