HackerRank Repeated String Problem Solution Yashwant Parihar, April 17, 2023April 18, 2023 In this post, we will solve HackerRank Repeated String Problem Solution. 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 ʼn letters of the infinite string.Examples = ‘abcac’n = 10The 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 Returns 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. Sample Input Sample Input 0 aba 10 Sample Output 0 7 Explanation 0The first n = 10 letters of the infinite string are abaabaabaa. Because there are 7 a’s, we return 7. Sample Input 1 a 1000000000000 Sample Output 1 1000000000000 Explanation 1Because all of the first n = 1000000000000 letters of the infinite string are a, we return1000000000000. HackerRank Repeated String Problem Solution Repeated String C Solution #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)); scanf("%s",s); long n; scanf("%ld",&n); unsigned long long int cnt = 0; for( unsigned long long int i = 0; s[i] != '\0'; i++) { if( s[i] == 'a') cnt++; } unsigned long long int cnt1 = 0; for( unsigned long long int i = 0; i < n%strlen(s); i++) { if( s[i] == 'a') cnt1++; } printf("%llu", cnt* (n/strlen(s)) + cnt1); return 0; } Repeated String C++ Solution #include <cmath> #include <cstdio> #include <vector> #include <iostream> #include <algorithm> using namespace std; int main() { /* Enter your code here. Read input from STDIN. Print output to STDOUT */ string s; cin>>s; long long n; cin>>n; int count=0; for(int i=0;i<s.length();i++){ if(s[i]=='a'){ count++; } } if(count==0){ cout<<'0'<<endl; } else if(n%s.length()==0){ cout<<(n/s.length())*count<<endl; } else{ long long temp=n/s.length()*count; for(int i=0;i<n%s.length();i++){ if(s[i]=='a'){ temp++; } } cout<<temp; } return 0; } Repeated String C Sharp Solution using System; using System.Collections.Generic; using System.IO; using System.Linq; class Solution { static void Main(String[] args) { string s = Console.ReadLine(); long n = Convert.ToInt64(Console.ReadLine()); long count = 0; for (int i = 0; i < s.Length; i++) if (s[i] == 'a') count++; long amount = (n / s.Length) * count; int rem = Convert.ToInt32(n % s.Length); for (int i = 0; i < rem; i++) if (s[i] == 'a') amount++; Console.WriteLine(amount); } } Repeated String Java Solution import java.io.*; import java.math.*; import java.security.*; import java.text.*; import java.util.*; import java.util.concurrent.*; import java.util.function.*; import java.util.regex.*; import java.util.stream.*; import static java.util.stream.Collectors.joining; import static java.util.stream.Collectors.toList; class Result { /* * Complete the 'repeatedString' function below. * * The function is expected to return a LONG_INTEGER. * The function accepts following parameters: * 1. STRING s * 2. LONG_INTEGER n */ public static long repeatedString(String s, long n) { int[] moduloFrequency = new int[s.length()]; int count = 0; for( int i = 0; i < s.length(); i++) { if(s.charAt(i) == 'a') count++; moduloFrequency[i] = count; } if(n <= s.length()) return moduloFrequency[(int)(n) - 1]; int modulo = (int) (n % s.length()) == 0 ? 0 : moduloFrequency[(int) (n % s.length()) - 1]; long howManyFullTimes = (n / s.length()); return howManyFullTimes * count + modulo; } } public class Solution { public static void main(String[] args) throws IOException { BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in)); BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH"))); String s = bufferedReader.readLine(); long n = Long.parseLong(bufferedReader.readLine().trim()); long result = Result.repeatedString(s, n); bufferedWriter.write(String.valueOf(result)); bufferedWriter.newLine(); bufferedReader.close(); bufferedWriter.close(); } } Repeated String JavaScript Solution process.stdin.resume(); process.stdin.setEncoding('ascii'); var input_stdin = ""; var input_stdin_array = ""; var input_currentline = 0; process.stdin.on('data', function (data) { input_stdin += data; }); process.stdin.on('end', function () { input_stdin_array = input_stdin.split("\n"); main(); }); function readLine() { return input_stdin_array[input_currentline++]; } /////////////// ignore above this line //////////////////// function main() { var s = readLine(); var n = parseInt(readLine()); var subStrLength = s.length; fullStrCount = (s.match(/a/g) || []).length; subStrCount = (s.substring(0, n%subStrLength).match(/a/g) || []).length; //n%subStrLength console.log((Math.floor(n/subStrLength) * fullStrCount) + subStrCount); } Repeated String Python Solution #!/bin/python3 import sys s = input().strip() n = int(input().strip()) num_of_as = s.count('a') total = 0 if len(s) <= n: complete = n//len(s) rest_of_s = s[:(n%len(s))] total += ((num_of_as * complete) + rest_of_s.count('a')) else: total += s[:n].count('a') print(total) Other Solutions HackerRank Jumping on the Clouds Solution HackerRank Equalize the Array Problem Solution c C# C++ HackerRank Solutions java javascript python CcppCSharpHackerrank Solutionsjavajavascriptpython