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

Repeated String

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3730 Discussions

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  • basimneermunda
     + 0 comments

    Here is my Java 7 solution:

        // To count a's in string s
    int aCount = 0;
    // To count a's in infinite string
    long Totalcount = 0;

    // Counting a's in string s
    for (int i=0; i<s.length(); i++) {
        char letter = s.charAt(i);
        if (letter == 'a') { aCount++; }
    }

    // Counting a's in infinite string

    // Check whether n is a multiple of s.length()
    if (n % s.length() == 0) {
        Totalcount = (n / s.length()) * aCount;
    } 
    else {
        // Calculate the nearest multiple of s.length() to n
        long nearestMultiple = n - (n%s.length());
        // Add a's count up to nearest multiple
        Totalcount = (nearestMultiple / s.length()) * aCount;

        // Add the remaining a's count in the remaining string
        for (int i=0; i<n%s.length(); i++) {
            char letter = s.charAt(i);
            if (letter == 'a') { Totalcount++; }
        }
    }

    return Totalcount;

  • manaya
     + 0 comments

    Solution in TypeScript:

    function repeatedString(s: string, n: number): number {
    const sLen = s.length;
    // use regex to calculate the count of a's in s
    const aCount = (s.match(/a/g) || []).length;
    // how many times is the lenght of s in n
    const repCount = Math.trunc(n / sLen);
    // how many characters needs to be added so the repeated string's length matches n
    const remainder = n % sLen;
    // get a substring with a length equal to the remainder number
    const remainderSubstring = s.substring(0, remainder);
    // how many a's are present in the substring
    const aCountInSubstring =  (remainderSubstring.match(/a/g) || []).length;
    // result is the count of a's in the times needed to be repeated + the ones in the remainder substring
    return repCount * aCount + aCountInSubstring;
}

  • JanekPython09
     + 0 comments

    Easy python sol:

    def repeatedString(s, n):
    c = s.count("a")
    l = len(s)
    return c*(n//l) + s[:n%l].count("a")

  • bsnagarjun90
     + 0 comments

    Python solution

    def repeatedString(s, n):
    # Write your code here
    number_of_times_to_multiply = n // len(s)

    a_count = sum([1 if i == 'a' else 0 for i in s])

    res = a_count * number_of_times_to_multiply

    diff = n - (len(s) * number_of_times_to_multiply)

    new_s = s[:diff]

    a_count = sum([1 if i == 'a' else 0 for i in new_s])
    
    res = res + a_count
    
    return res

  • greivin_lopez
     + 0 comments
    def repeated_string(s, n):
    size = len(s)
    return n // size * list(s).count("a") + list(s)[:n % size].count("a")