What is the time complexity of Check if All Characters Have Equal Number of Occurrences?

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

What data structures are used to solve Check if All Characters Have Equal Number of Occurrences?

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 Check if All Characters Have Equal Number of Occurrences?

Insight 1: Use a hash table (or array) to count the occurrences of each character. This allows for efficient counting in O(n) time. Insight 2: Once character counts are tallied, iterate through them to check if all counts are equal. A simple variable to track the first encountered count simplifies the comparison logic.

Which companies ask LeetCode 1941. Check if All Characters Have Equal Number of Occurrences in interviews?

This problem is frequently asked at Bolt.

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Check if All Characters Have Equal Number of Occurrences - LeetCode 1941 Solution

Check if All Characters Have Equal Number of Occurrences - Complete Solution Guide

Check if All Characters Have Equal Number of Occurrences is LeetCode problem 1941, 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

Given a string s , return true if s is a good string, or false otherwise . A string s is good if all the characters that appear in s have the same number of occurrences (i.e., the same frequency). Example 1: Input: s = "abacbc" Output: true Explanation: The characters that appear in s are 'a', 'b', and 'c'. All characters occur 2 times in s. Example 2: Input: s = "aaabb" Output: false Explanation: The characters that appear in s are 'a' and 'b'. 'a' occurs 3 times while 'b' occurs 2 times, which

Detailed Explanation

The problem asks whether all characters in a given string have the same frequency of occurrence. The input is a string `s` containing lowercase English letters. The output is a boolean value: `true` if all characters present in the string appear the same number of times, and `false` otherwise. The string's length is constrained between 1 and 1000 characters.

Solution Approach

The provided solutions all use a frequency counting approach. First, they iterate through the input string `s`, counting the occurrences of each character. In the Python solution, a dictionary is used; in Java, C++, and C, an array of size 26 is used to store the frequencies (one for each letter 'a' to 'z'). After counting, the solutions iterate through the counts. If a count is different from the first non-zero count found, the function immediately returns `false`. Otherwise, if all counts are equal or only one character was present, the function returns `true`.

Step-by-Step Algorithm

Step 1: Initialize a data structure (dictionary or array of size 26) to store character frequencies. Initialize all counts to 0.
Step 2: Iterate through the input string `s`. For each character, increment its corresponding frequency count in the data structure.
Step 3: Iterate through the frequency counts. If this is the first non-zero count encountered, store it. If any subsequent non-zero count differs from the stored count, return `false`.
Step 4: If all non-zero counts are equal to the first encountered count, return `true`.

Key Insights

Insight 1: Use a hash table (or array) to count the occurrences of each character. This allows for efficient counting in O(n) time.
Insight 2: Once character counts are tallied, iterate through them to check if all counts are equal. A simple variable to track the first encountered count simplifies the comparison logic.
Insight 3: The space complexity can be optimized to O(1) because we only need to store counts for 26 lowercase English letters. An array of size 26 is sufficient instead of a general hash map.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Hash Table, String, Counting.

Companies

Asked at: Bolt.

Check if All Characters Have Equal Number of Occurrences | LeetCode 1941

Easy

Hash Table String Counting

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public boolean areOccurrencesEqual(String s) {
        int[] freq = new int[26];
        for (char c : s.toCharArray()) {
            freq[c - 'a']++;
        }

        int count = 0;
        for (int i = 0; i < 26; i++) {
            if (freq[i] > 0) {
                if (count == 0) {
                    count = freq[i];
                } else if (freq[i] != count) {
                    return false;
                }
            }
        }

        return true;
    }
}

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Problem Info

Description

Problem 1941: Check if All Characters Have Equal Number of Occurrences - Detailed explanation and solution approaches.

Topics

Hash Table String Counting

Companies

Bolt

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1941

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

4 steps

1
Step 1: Initialize a data structure (dictionary or array of size 26) to store character frequencies. Initialize all counts to 0.
2
Step 2: Iterate through the input string `s`. For each character, increment its corresponding frequency count in the data structure.
3
Step 3: Iterate through the frequency counts. If this is the first non-zero count encountered, store it. If any subsequent non-zero count differs from the stored count, return `false`.
4
Step 4: If all non-zero counts are equal to the first encountered count, 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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The first loop iterates through the string once (O(n)). The second loop iterates through the frequency array which has a fixed size of 26 (O(1)). Therefore, the dominant factor is the linear traversal of the string, leading to O(n) time complexity. The space used is only the frequency array, which is a constant size regardless of the input string length, leading to O(1) space complexity.

📖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 `Counter` object (Python)

Python's `collections.Counter` object can directly count character frequencies, potentially simplifying the code.

Time: O(n)Space: O(1)

Trade-offs: Slightly more concise code, but adds a dependency on the `collections` module. Performance is comparable.

✨

Key Insights

1Insight 1: Use a hash table (or array) to count the occurrences of each character. This allows for efficient counting in O(n) time.
2Insight 2: Once character counts are tallied, iterate through them to check if all counts are equal. A simple variable to track the first encountered count simplifies the comparison logic.
3Insight 3: The space complexity can be optimized to O(1) because we only need to store counts for 26 lowercase English letters. An array of size 26 is sufficient instead of a general hash map.

📚

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 string. The solutions gracefully handle this case either implicitly (by not entering the loops if the string is empty) or explicitly (Python handles this implicitly).
Edge case 2: String with only one unique character. The solutions correctly identify this as a 'good' string because all characters (only one) have the same frequency.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle edge cases, particularly empty strings. The code needs to check for this condition and act accordingly.

✗Common mistake 2: Inefficiently checking for equal frequencies. A straightforward comparison with the first observed count improves efficiency over comparing all counts to each other.

✅

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 validation. This algorithm could be used to validate data entries where the expected pattern involves all characters having the same frequency (e.g., certain types of cryptographic keys or codes).

🏦Real-world scenario 2: Text analysis. It can be used in linguistic analysis to identify unusual word patterns or characteristics in text where an unexpected character frequency might indicate an error or anomaly.

💡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

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

⚠️ 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: October 20, 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

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

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

@InterviewProJul 7

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

@CodeNinjaJun 30

❓

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 "Check if All Characters Have Equal Number of Occurrences"?

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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