String Alphabet Count

Count the frequency of each alphabet character in a given string and display them in alphabetical order.

Understand the Problem

Problem Statement

String Alphabet Count: Given a string S with only alphabets, print the alphabet and its count as shown in the Example Input/Output section.

Constraints

1 <= Length of S <= 100
String S contains only alphabetic characters (a-z, A-Z)
Case sensitivity: treat uppercase and lowercase as distinct characters
Output must be sorted alphabetically by character

Examples

Example 1

Input

apple

Output

a1 e1 l1 p2

Explanation

In 'apple': 'a' appears 1 time, 'e' appears 1 time, 'l' appears 1 time, 'p' appears 2 times. Output is sorted alphabetically.

Example 2

Input

hello

Output

e1 h1 l2 o1

Explanation

In 'hello': 'e' appears 1 time, 'h' appears 1 time, 'l' appears 2 times, 'o' appears 1 time. Output is sorted alphabetically.

Example 3

Input

aabbcc

Output

a2 b2 c2

Explanation

In 'aabbcc': each character 'a', 'b', and 'c' appears exactly 2 times. Output is sorted alphabetically.

Solution

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

int compare(const void *a, const void *b) {
    return (*(char*)a - *(char*)b);
}

int main() {
    char s[101];
    fgets(s, sizeof(s), stdin);
    
    // Remove newline if present
    int len = strlen(s);
    if (s[len-1] == '\n') {
        s[len-1] = '\0';
        len--;
    }
    
    // Create sorted copy for ordering
    char sorted[101];
    strcpy(sorted, s);
    qsort(sorted, len, sizeof(char), compare);
    
    // Create unique character array
    char unique[101];
    int unique_count = 0;
    
    for (int i = 0; i < len; i++) {
        int found = 0;
        for (int j = 0; j < unique_count; j++) {
            if (unique[j] == sorted[i]) {
                found = 1;
                break;
            }
        }
        if (!found) {
            unique[unique_count++] = sorted[i];
        }
    }
    
    // Print results
    for (int i = 0; i < unique_count; i++) {
        int count = 0;
        for (int j = 0; j < len; j++) {
            if (s[j] == unique[i]) {
                count++;
            }
        }
        printf("%c%d", unique[i], count);
        if (i < unique_count - 1) {
            printf(" ");
        }
    }
    
    return 0;
}

Time:O(n log n) due to sorting

Space:O(n) for storing sorted and unique character arrays

Approach:

C Solution Explanation:

1. Read input string using fgets() and remove trailing newline

2. Create a sorted copy of the string using qsort() to determine output order

3. Extract unique characters from sorted string using nested loops

4. For each unique character, count occurrences in original string

5. Print character and count, with spaces between pairs

6. Uses standard C library functions for sorting and string manipulation

Visual Explanation

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