Check if All Characters Have Equal Number of Occurrences - Problem

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).

Input & Output

Example 1 — Good String

$ Input: s = "abacbc"

› Output: true

💡 Note: Each character appears exactly 2 times: 'a' appears 2 times, 'b' appears 2 times, 'c' appears 2 times. All frequencies are equal.

Example 2 — Bad String

$ Input: s = "aaabb"

› Output: false

💡 Note: Character 'a' appears 3 times but 'b' appears 2 times. The frequencies are not equal.

Example 3 — Single Character

$ Input: s = "c"

› Output: true

💡 Note: Only one unique character 'c' appears 1 time. All characters have the same frequency (trivially true).

Constraints

1 ≤ s.length ≤ 1000
s consists of lowercase English letters

Visualization

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Asked in

a Amazon 15 M Microsoft 12

The key insight is to count character frequencies and verify they are all equal. The hash map approach is optimal: count all frequencies in O(n) time, then check equality. Time: O(n), Space: O(k) where k is unique characters.

Common Approaches

✓ Brute Force Character Counting

⏱️ Time: O(n²) Space: O(1)

For each unique character in the string, count its occurrences by scanning the entire string. Then compare all counts to ensure they are equal.

Hash Map Frequency Count

⏱️ Time: O(n) Space: O(k)

Count the frequency of each character using a hash map in a single pass, then check if all frequencies are equal.

Brute Force Character Counting — Algorithm Steps

Find all unique characters
Count each character's frequency separately
Compare all frequencies to check equality

Visualization

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Step-by-Step Walkthrough

Find Unique Characters

Identify all unique characters: a, b, c

Count Each Character

Count 'a': 2, 'b': 2, 'c': 2

Compare Frequencies

All frequencies are equal (2), so return true

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>

bool solution(char* s) {
    int len = strlen(s);
    if (len == 0) return true;
    
    // Find unique characters
    char uniqueChars[256];
    int uniqueCount = 0;
    
    for (int i = 0; i < len; i++) {
        bool found = false;
        for (int j = 0; j < uniqueCount; j++) {
            if (uniqueChars[j] == s[i]) {
                found = true;
                break;
            }
        }
        if (!found) {
            uniqueChars[uniqueCount++] = s[i];
        }
    }
    
    // Count frequency of first character
    int firstCount = 0;
    for (int i = 0; i < len; i++) {
        if (s[i] == uniqueChars[0]) {
            firstCount++;
        }
    }
    
    // Check if all other characters have same frequency
    for (int i = 1; i < uniqueCount; i++) {
        int count = 0;
        for (int j = 0; j < len; j++) {
            if (s[j] == uniqueChars[i]) {
                count++;
            }
        }
        if (count != firstCount) {
            return false;
        }
    }
    
    return true;
}

int main() {
    char s[1001];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    
    bool result = solution(s);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each unique character, we scan the entire string

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables to store counts and comparisons

✓ Linear Space

28.4K Views

Medium Frequency

~10 min Avg. Time

856 Likes

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Check if All Characters Have Equal Number of Occurrences - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Character Counting — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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