Character Frequency Counter - Problem

Given a string, find the frequency of each character and display the results in descending order of frequency.

If two characters have the same frequency, they can appear in any order relative to each other.

Input: A single string containing letters, numbers, and special characters

Output: A string showing each character and its frequency, sorted by frequency in descending order

Format: Each character-frequency pair should be displayed as "character:frequency", separated by commas

Input & Output

Example 1 — Basic Case

$ Input: s = "hello"

› Output: l:2,h:1,e:1,o:1

💡 Note: Character frequencies: h→1, e→1, l→2, o→1. Sorted by frequency descending: l appears 2 times (highest), then h, e, o each appear 1 time

Example 2 — Mixed Characters

$ Input: s = "aabbcc"

› Output: a:2,b:2,c:2

💡 Note: All characters have same frequency (2), so they can appear in any order relative to each other

Example 3 — Single Character

$ Input: s = "aaaa"

› Output: a:4

💡 Note: Only one unique character 'a' appears 4 times

Constraints

1 ≤ s.length ≤ 1000
s contains only printable ASCII characters

Visualization

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Asked in

G Google 25 a Amazon 20 M Microsoft 15 🍎 Apple 10

The key insight is to use a hash map to count character frequencies in a single pass through the string, then sort by frequency. The optimal approach uses frequency counting with hash map - O(n) counting + O(k log k) sorting. Time: O(n + k log k), Space: O(k) where k is number of unique characters.

Common Approaches

✓ Brute Force - Count Each Character Separately

⏱️ Time: O(n²) Space: O(k)

Find all unique characters first, then for each unique character, iterate through the entire string counting its occurrences. Finally, sort by frequency in descending order.

Hash Map Frequency Counter

⏱️ Time: O(n + k log k) Space: O(k)

Iterate through the string once, building a frequency map. Then convert the map to a list and sort by frequency in descending order.

Brute Force - Count Each Character Separately — Algorithm Steps

Find all unique characters in the string
For each unique character, count its occurrences by scanning entire string
Create frequency pairs and sort by count descending
Format output as character:frequency pairs

Visualization

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Step-by-Step Walkthrough

Find Unique Characters

Extract all unique characters from string

Count Each Character

For each unique char, scan entire string counting occurrences

Sort & Format

Sort by frequency descending and format output

Code -

solution.c — C

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

typedef struct {
    char ch;
    int freq;
} CharFreq;

int compare(const void *a, const void *b) {
    CharFreq *cf1 = (CharFreq *)a;
    CharFreq *cf2 = (CharFreq *)b;
    return cf2->freq - cf1->freq;
}

char* solution(char *s) {
    int len = strlen(s);
    if (len == 0) {
        char *result = malloc(1);
        result[0] = '\0';
        return result;
    }
    
    // Find unique characters
    char unique[256];
    int uniqueCount = 0;
    int seen[256] = {0};
    
    for (int i = 0; i < len; i++) {
        if (!seen[(unsigned char)s[i]]) {
            seen[(unsigned char)s[i]] = 1;
            unique[uniqueCount++] = s[i];
        }
    }
    
    // Count frequency for each unique character
    CharFreq freqPairs[256];
    for (int i = 0; i < uniqueCount; i++) {
        int count = 0;
        for (int j = 0; j < len; j++) {
            if (s[j] == unique[i]) {
                count++;
            }
        }
        freqPairs[i].ch = unique[i];
        freqPairs[i].freq = count;
    }
    
    // Sort by frequency descending
    qsort(freqPairs, uniqueCount, sizeof(CharFreq), compare);
    
    // Format output
    char *result = malloc(len * 10);  // Generous allocation
    result[0] = '\0';
    
    for (int i = 0; i < uniqueCount; i++) {
        if (i > 0) strcat(result, ",");
        char temp[20];
        sprintf(temp, "%c:%d", freqPairs[i].ch, freqPairs[i].freq);
        strcat(result, temp);
    }
    
    return result;
}

int main() {
    char s[1001];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    
    char *result = solution(s);
    printf("%s\n", result);
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each unique character, we scan the entire string, leading to nested loops

⚠ Quadratic Growth

Space Complexity

O(k)

Space needed to store k unique characters and their frequencies

✓ Linear Space

25.0K Views

Medium Frequency

~15 min Avg. Time

850 Likes

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Character Frequency Counter - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Count Each Character Separately — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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