What is the time complexity of Redistribute Characters to Make All Strings Equal?

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

What data structures are used to solve Redistribute Characters to Make All Strings Equal?

This problem uses Hash Table, String, Counting. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Redistribute Characters to Make All Strings Equal?

Insight 1: The problem boils down to checking if the frequency of each character across all strings is divisible by the number of strings. If it is, you can redistribute the characters evenly. Insight 2: Using a hash table (or array) to count character frequencies provides an efficient way to track character distribution.

Which companies ask LeetCode 1897. Redistribute Characters to Make All Strings Equal in interviews?

This problem is frequently asked at Moengage.

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Redistribute Characters to Make All Strings Equal - LeetCode 1897 Solution

Redistribute Characters to Make All Strings Equal - Complete Solution Guide

Redistribute Characters to Make All Strings Equal is LeetCode problem 1897, 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 an array of strings words ( 0-indexed ). In one operation, pick two distinct indices i and j , where words[i] is a non-empty string, and move any character from words[i] to any position in words[j] . Return true if you can make every string in words equal using any number of operations , and false otherwise . Example 1: Input: words = ["abc","aabc","bc"] Output: true Explanation: Move the first 'a' in words[1] to the front of words[2], to make words[1] = "abc" and words[2] = "abc".

Detailed Explanation

The problem asks whether it's possible to make all strings in an input array `words` equal by moving characters between them. The operation allows moving any character from one string to another. The goal is to determine if it's possible to achieve a state where all strings are identical, regardless of their final length or content. The input is an array of strings, and the output is a boolean: `true` if it's possible to make all strings equal, and `false` otherwise.

Solution Approach

The solution uses a frequency counting approach. It iterates through all strings in the `words` array. For each character in each string, it increments the count of that character in a hash table (or array). After counting all characters, it checks if the frequency of each character is a multiple of the number of strings. If any character's frequency is not divisible by the number of strings, it means equal distribution is impossible, so the function returns `false`. Otherwise, it returns `true`.

Step-by-Step Algorithm

Step 1: Initialize a hash table (or array) `counts` to store the frequency of each character (a-z). Initialize all counts to 0.
Step 2: Iterate through each string in the `words` array.
Step 3: Iterate through each character in the current string.
Step 4: Increment the count of the current character in the `counts` hash table.
Step 5: After processing all strings, iterate through the `counts` hash table.
Step 6: For each character's count, check if it's divisible by the number of strings (`words.length`).
Step 7: If any count is not divisible by the number of strings, return `false`. Otherwise, return `true`.

Key Insights

Insight 1: The problem boils down to checking if the frequency of each character across all strings is divisible by the number of strings. If it is, you can redistribute the characters evenly.
Insight 2: Using a hash table (or array) to count character frequencies provides an efficient way to track character distribution.
Insight 3: No need to actually perform the character redistribution. The solution focuses solely on verifying the divisibility condition for each character's frequency.

Complexity Analysis

Time Complexity: O(N)

Space Complexity: O(1)

Topics

This problem involves: Hash Table, String, Counting.

Companies

Asked at: Moengage.

Redistribute Characters to Make All Strings Equal | LeetCode 1897

Easy

Hash Table String Counting

Time: O(N)Space: O(1)

Solution Code

Available in:

class Solution {
    public boolean makeEqual(String[] words) {
        int[] charCounts = new int[26];
        for (String word : words) {
            for (char c : word.toCharArray()) {
                charCounts[c - 'a']++;
            }
        }

        int numWords = words.length;
        for (int count : charCounts) {
            if (count % numWords != 0) {
                return false;
            }
        }

        return true;
    }
}

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Problem Info

Description

Problem 1897: Redistribute Characters to Make All Strings Equal - Detailed explanation and solution approaches.

Topics

Hash Table String Counting

Companies

Moengage

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1897

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 hash table (or array) `counts` to store the frequency of each character (a-z). Initialize all counts to 0.
2
Step 2: Iterate through each string in the `words` array.
3
Step 3: Iterate through each character in the current string.
4
Step 4: Increment the count of the current character in the `counts` hash table.
5
Step 5: After processing all strings, iterate through the `counts` hash table.
6
Step 6: For each character's count, check if it's divisible by the number of strings (`words.length`).
7
Step 7: If any count is not divisible by the number of strings, return `false`. Otherwise, return `true`.

💡

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)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The nested loops dominate the time complexity. The space used is independent of the input size, only dependent on the fixed size of the alphabet.

📖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

Using a different data structure

Instead of a hash table (or array), one could use a map (in Python or Java) or unordered map (in C++) to store character counts. This would offer more flexibility if dealing with a wider character set. However, this adds slight overhead for lookup compared to direct array indexing.

Time: O(N*M)Space: O(K), where K is the number of unique characters (worst case O(M))

Trade-offs: Using a more flexible data structure increases space complexity in worst-case scenarios, but might be necessary if dealing with larger character sets or non-ASCII characters.

✨

Key Insights

1Insight 1: The problem boils down to checking if the frequency of each character across all strings is divisible by the number of strings. If it is, you can redistribute the characters evenly.
2Insight 2: Using a hash table (or array) to count character frequencies provides an efficient way to track character distribution.
3Insight 3: No need to actually perform the character redistribution. The solution focuses solely on verifying the divisibility condition for each character's frequency.

📚

Key Terms Explained

For Students

Hash Table

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

String

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

Edge Cases

Edge case 1: Empty input array. The solution correctly handles this as it would iterate zero times and return true (implicitly).
Edge case 2: An array with strings of different lengths. The solution still works as it counts character frequencies regardless of string length.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check all character frequencies. The solution must check divisibility for *every* character.

✗Common mistake 2: Incorrectly calculating or using the number of strings. The divisibility check must always be against the total number of strings in the input.

✅

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: Resource allocation. Determining if a set of resources can be evenly distributed among several tasks or users.

🏦Real-world scenario 2: Data balancing in distributed systems. Checking if data can be evenly spread across multiple servers or nodes.

💡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

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ 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: 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 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.

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 "Redistribute Characters to Make All Strings Equal"?

For this easy-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.

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