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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Repeated String
  5. Discussions

Repeated String

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3720 Discussions

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  • pallavisangem23
     + 0 comments
    public static long repeatedString(String s, long n) {
    // Write your code here
    long count=0;
      for(int i=0;i<s.length();i++){
        if(s.charAt(i)=='a') count++;
      }
      count=count*(n/s.length());
      long rem=(n%(s.length()));
      for(int j=0;j<rem;j++){
         if(s.charAt(j)=='a') count++;
      }
     return count;
    }

  • giorgibregvadze1
     + 0 comments

    Here's a Javascript solution:

    function repeatedString(s, n) {
    const remaining = Math.round((n / s.length - Math.floor(n / s.length)) * s.length);
    const aInSingle = (s.match(/a/g) || []).length;
    return Math.floor(n/s.length)*aInSingle + (s.substring(0,remaining).match(/a/g) || []).length
}

  • yashdeora98294
     + 0 comments

    Here is problem solution in Python, java, C++, C and Javascript - https://programmingoneonone.com/hackerrank-repeated-string-problem-solution.html

  • akrishna34
     + 0 comments

    def repeatedString(s, n):

    count=0
count=count_a(s)
if(n%len(s)==0):
    count=count*(n//len(s))
elif(n%len(s)!=0):
    count=count*(n//len(s))+count_a(s[0:(n%len(s))])
return count

    def count_a(s): count=0 for i in s: if (i=="a"): count+=1 return count

  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can watch the explanation here : https://youtu.be/Vh5davsSkfA

    long ccount(string s) {
    return count(s.begin(), s.end(), 'a');
}

long repeatedString(string s, long n) {
    return (n / s.size()) * ccount(s) + ccount(s.substr(0, n % s.size()));
}