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Alphabet Count – Repeated String

Count occurrences of a specific alphabet in a string that has been repeated to reach a specified length

Understand the Problem

Problem Statement

A string S is passed as input to the program. The string S is repeated till the repeated string R is of length N. The program must print the count of a specific alphabet A which is passed as the input in the repeated string R.

Constraints

  • 1 <= Length of S <= 50
  • 1 <= N <= 9999999
  • A is from a to z

Examples

Example 1
Input
abcd
10
b
Output
3
Explanation

abcd when repeated till length 10 is abcdabcdab in which the alphabet b occurs 3 times. The string 'abcd' has 1 occurrence of 'b'. We have 2 complete repetitions (8 characters) plus 2 more characters 'ab'. Total = 2×1 + 1 = 3.

Solution

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

int main() {
    char s[51];
    long long N;
    char A;
    
    // Read inputs
    scanf("%s", s);
    scanf("%lld", &N);
    scanf(" %c", &A);  // Space before %c to consume newline
    
    int len = strlen(s);
    
    // Count occurrences of A in original string
    int count_in_original = 0;
    for (int i = 0; i < len; i++) {
        if (s[i] == A) {
            count_in_original++;
        }
    }
    
    // Calculate complete repetitions and remainder
    long long complete_repetitions = N / len;
    int remainder = N % len;
    
    // Count in complete repetitions
    long long total_count = complete_repetitions * count_in_original;
    
    // Count in partial repetition (if any)
    for (int i = 0; i < remainder; i++) {
        if (s[i] == A) {
            total_count++;
        }
    }
    
    printf("%lld", total_count);
    
    return 0;
}
Time:O(len(S)) - We only iterate through the original string twice: once to count occurrences and once for the remainder
Space:O(len(S)) - We store the original string, which is at most 50 characters
Approach:

C Solution Logic:

  1. Read the string S, target length N, and alphabet A
  2. Calculate length of original string
  3. Count how many times A appears in the original string by iterating through it
  4. Calculate complete repetitions using integer division: N / len(S)
    5. <5>Calculate remainder characters using modulo: N % len(S)<6>Count total occurrences: complete_repetitions × count_in_original<7>For remaining characters, count occurrences in the prefix of original string<8>Output the final count

Visual Explanation

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