Repeated String

Problem Statement :

There is a string, s, of lowercase English letters that is repeated infinitely many times. Given an integer, n, find and print the number of letter a's in the first n letters of the infinite string.


s = 'abcac'
n = 10

The substring we consider is abcacabcac, the first 10 characters of the infinite string. There are  4 occurrences of a in the substring.

Function Description

Complete the repeatedString function in the editor below.

repeatedString has the following parameter(s):

s: a string to repeat
n: the number of characters to consider

int: the frequency of a in the substring
Input Format

The first line contains a single string, s.
The second line contains an integer, n.


1  <=   | s |  <=  100
1  <=    n     <=   10^12

For 25%  of the test cases, n <= 10^6 .

Solution :


                            Solution in C :

In    C :

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

int main(){
    char* s = (char *)malloc(512000 * sizeof(char));
    long n,o,p,i; 
    return 0;

                        Solution in C++ :

In    C ++ :

#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <set>
#include <string>
#include <map>
#include <ctime>
#include <cstdlib>
#include <unordered_set>

using namespace std;

typedef long long ll;

const int MaxN = 100;

char s[MaxN + 1];

int main() {
    ll n;
    scanf ("%s", s);
    int len = (int)strlen(s);
    scanf ("%lld", &n);
    ll tot = n / len;
    ll as = 0;
    for (int i = 0; i < len; ++i)
        if (s[i] == 'a')
    ll res = tot * as;
    for (int i = 0; i < n % len; ++i) 
        if (s[i] == 'a')
    printf ("%lld", res);

                        Solution in Java :

In   Java  :

import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(;
        String s =;
        long n = in.nextLong();
        long num = n/s.length();
        long rem = n%s.length();
        long ans = 0;
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i)=='a') {
                ans += num;
                if (i < rem)

                        Solution in Python : 
In   Python3  :


import math
import os
import random
import re
import sys

# Complete the repeatedString function below.
def repeatedString(s, n):
    return (s.count('a') * (n//len(s))+s[:n%len(s)].count('a'))

if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    s = input()

    n = int(input())

    result = repeatedString(s, n)

    fptr.write(str(result) + '\n')


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