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

Repeated String

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3717 Discussions

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  • 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()));
}

  • nokors
     + 0 comments

    Python full_repeats = n // len(s) remainder = n % len(s) a_count = s.count('a') total_a_count = full_repeats * a_count total_a_count += s[:remainder].count('a') return total_a_count `

  • esilva3
     + 0 comments

    My python3 solution

    l = len(s)

return( s[:n%l].count('a') + (n//l * s.count('a')))

  • saurabh_srivast2
     + 0 comments
    def repeatedString(s, n):
    # Write your code here
    slen =  len(s)
    count1 = s.count('a')
    n1 = n // slen
    n2 = n % slen
    return (count1 * n1 ) + s.count('a',0, n2)