What is the time complexity of Reorganize String?

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

What data structures are used to solve Reorganize String?

This problem uses Hash Table, String, Greedy, Sorting, Heap (Priority Queue), Counting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Reorganize String?

Insight 1: The most frequent character in the string can appear at most (n + 1) / 2 times, where n is the length of the string. Otherwise, it's impossible to rearrange the string without adjacent characters being the same. Insight 2: Using a priority queue (or max-heap) to keep track of the most frequent characters is efficient for placing characters in the rearranged string.

Which companies ask LeetCode 767. Reorganize String in interviews?

This problem is frequently asked at Citadel, Coupang, Druva, Intuit, PayPal and 5 more companies.

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

Reorganize String - LeetCode 767 Solution

Q: How to solve Reorganize String?

The solution uses a greedy approach combined with a priority queue (or sorting in the C implementation). First, it counts the frequency of each character in the input string. If any character appears more than (n + 1) / 2 times, it's impossible to rearrange, so it returns an empty string. Otherwise, it uses a priority queue (or sorts) to store characters and their counts, ordered by frequency. It then iterates, adding the most frequent character at the next available index in the `ans` array. It prioritizes placing characters at even indices first (0, 2, 4, ...). If the end of the array is reached, it starts again at index 1 to continue filling the array. The resulting character array is then converted into a string and returned.

Reorganize String - Complete Solution Guide

Reorganize String is LeetCode problem 767, 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 , rearrange the characters of s so that any two adjacent characters are not the same. Return any possible rearrangement of s or return "" if not possible . Example 1: Input: s = "aab" Output: "aba" Example 2: Input: s = "aaab" Output: "" Constraints: 1 <= s.length <= 500 s consists of lowercase English letters.

Detailed Explanation

The problem asks us to rearrange the characters of a given string 's' such that no two adjacent characters are the same. The goal is to return any possible rearrangement that satisfies this condition. If it's impossible to rearrange the string in this way, we should return an empty string.

Solution Approach

The solution uses a greedy approach combined with a priority queue (or sorting in the C implementation). First, it counts the frequency of each character in the input string. If any character appears more than (n + 1) / 2 times, it's impossible to rearrange, so it returns an empty string. Otherwise, it uses a priority queue (or sorts) to store characters and their counts, ordered by frequency. It then iterates, adding the most frequent character at the next available index in the `ans` array. It prioritizes placing characters at even indices first (0, 2, 4, ...). If the end of the array is reached, it starts again at index 1 to continue filling the array. The resulting character array is then converted into a string and returned.

Step-by-Step Algorithm

Step 1: Count the frequency of each character in the input string 's'.
Step 2: Check if any character's frequency is greater than (n + 1) / 2. If yes, return an empty string.
Step 3: Create a priority queue (or sort the character counts). The priority queue should store characters and their counts, ordered by the count in descending order.
Step 4: Create a character array 'ans' of the same size as the input string 's'.
Step 5: Iterate while the priority queue is not empty.
Step 6: For each character in the priority queue, place it into the 'ans' array. Start by placing it at index 'idx', then increment 'idx' by 2. If 'idx' goes beyond the length of the string, reset 'idx' to 1.
Step 7: Convert the character array 'ans' to a string and return it.

Key Insights

Insight 1: The most frequent character in the string can appear at most (n + 1) / 2 times, where n is the length of the string. Otherwise, it's impossible to rearrange the string without adjacent characters being the same.
Insight 2: Using a priority queue (or max-heap) to keep track of the most frequent characters is efficient for placing characters in the rearranged string.
Insight 3: Distributing the most frequent characters first and placing them at even indices, and then shifting to odd indices if necessary, helps in avoiding adjacent duplicates.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Hash Table, String, Greedy, Sorting, Heap (Priority Queue), Counting.

Companies

Asked at: Citadel, Coupang, Druva, Intuit, PayPal, Pinterest, Roblox, Tesla, Walmart Labs, eBay.

Reorganize String | LeetCode 767

Medium

Hash Table String Greedy Sorting Heap (Priority Queue)Counting

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

Solution Code

Available in:

import java.util.PriorityQueue;
import java.util.HashMap;
import java.util.Map;

class Solution {
    public String reorganizeString(String s) {
        int n = s.length();
        Map<Character, Integer> counts = new HashMap<>();
        for (char c : s.toCharArray()) {
            counts.put(c, counts.getOrDefault(c, 0) + 1);
        }

        for (int count : counts.values()) {
            if (count > (n + 1) / 2) {
                return "";
            }
        }

        char[] ans = new char[n];
        int idx = 0;

        PriorityQueue<Map.Entry<Character, Integer>> pq = new PriorityQueue<>((a, b) -> b.getValue() - a.getValue());
        pq.addAll(counts.entrySet());

        while (!pq.isEmpty()) {
            Map.Entry<Character, Integer> entry = pq.poll();
            char character = entry.getKey();
            int count = entry.getValue();

            for (int i = 0; i < count; i++) {
                ans[idx] = character;
                idx += 2;
                if (idx >= n) {
                    idx = 1;
                }
            }
        }

        return new String(ans);
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 767: Reorganize String - Detailed explanation and solution approaches.

Topics

Hash Table String Greedy Sorting Heap (Priority Queue)Counting

Companies

CitadelCoupangDruvaIntuitPayPalPinterestRobloxTesla

Stats

Difficulty:Medium

Languages:4

Companies:10

Problem #:767

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: Count the frequency of each character in the input string 's'.
2
Step 2: Check if any character's frequency is greater than (n + 1) / 2. If yes, return an empty string.
3
Step 3: Create a priority queue (or sort the character counts). The priority queue should store characters and their counts, ordered by the count in descending order.
4
Step 4: Create a character array 'ans' of the same size as the input string 's'.
5
Step 5: Iterate while the priority queue is not empty.
6
Step 6: For each character in the priority queue, place it into the 'ans' array. Start by placing it at index 'idx', then increment 'idx' by 2. If 'idx' goes beyond the length of the string, reset 'idx' to 1.
7
Step 7: Convert the character array 'ans' 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 time complexity is dominated by iterating through the string to count frequencies and building the result string. The priority queue operations are effectively constant time because the number of distinct characters is limited to 26. The space complexity is mainly due to storing the result 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

No priority queue

Instead of using a priority queue, one could sort the counts array after counting. It is slightly less efficient, but easier to understand.

Time: O(n + k log k), where k is the number of distinct characters (at most 26). The n comes from initial counts.Space: O(k) where k is number of distinct characters (at most 26)

Trade-offs: Slightly less efficient than priority queue, but simpler to implement for a small alphabet size.

✨

Key Insights

1Insight 1: The most frequent character in the string can appear at most (n + 1) / 2 times, where n is the length of the string. Otherwise, it's impossible to rearrange the string without adjacent characters being the same.
2Insight 2: Using a priority queue (or max-heap) to keep track of the most frequent characters is efficient for placing characters in the rearranged string.
3Insight 3: Distributing the most frequent characters first and placing them at even indices, and then shifting to odd indices if necessary, helps in avoiding adjacent duplicates.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Greedy Algorithm

Making the best choice at each step without looking ahead.

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: Input string is empty or has only one character. The algorithm handles these cases correctly as the condition count > (n+1)/2 will never be true, and the correct rearranged string (the string itself) will be returned.
Edge case 2: Input string is impossible to rearrange (e.g., 'aaab'). The algorithm correctly identifies this and returns an empty string because one of the character counts would exceed (n + 1) / 2.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check if a rearrangement is possible before attempting to rearrange. This leads to incorrect output for strings like 'aaab'.

✗Common mistake 2: Incorrectly calculating the maximum allowed frequency. Using n/2 instead of (n+1)/2 leads to errors.

✗Common mistake 3: Not handling the even/odd index placement correctly, leading to adjacent duplicates in the result.

✅

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: Load balancing: Distributing tasks across multiple servers such that no single server is overloaded, conceptually similar to distributing characters to avoid adjacent duplicates.

🏦Real-world scenario 2: Data interleaving: Interleaving data streams from different sources to improve read/write performance, similar to distributing characters to avoid repetitions.

💡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

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

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

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

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: July 18, 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 choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

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 "Reorganize String"?

For this medium-level Hash Table 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?