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

Repeated String

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3704 Discussions

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  • jayantnishad34
     + 0 comments

    public static long repeatedString(String s, long n) { // Write your code here long strLength = s.length();

    // Count occurrences of 'a' in the original string
long countInOriginal = s.chars().filter(ch -> ch == 'a').count();

// Calculate full repetitions of `s` in `n` characters
long fullRepetitions = n / strLength;

// Calculate remaining characters
long remainder = n % strLength;

// Count occurrences of 'a' in the remainder substring
long countInRemainder = s.substring(0, (int) remainder).chars().filter(ch -> ch == 'a').count();

// Total occurrences of 'a'
return (fullRepetitions * countInOriginal) + countInRemainder;
}

  • 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()));
}

  • gabriel_almilha1
     + 0 comments

    Python

    def repeatedString(string: str, substring_size: int)-> int:
    # This code can be simplified, but maintainability wins
    # How many characters are in the repeating string
    size: int = len(string)
    # If the number of times the string repeats has remainder, account it
    last_position: int = substring_size % size
    # Number of times the string repeats entirely
    full_repeats: int = substring_size // size
    
    return string.count('a') * full_repeats + string[0: last_position].count('a')

  • robertbielicki
     + 0 comments
    def repeatedString(s: str, n: int) -> int:
    return s.count('a') * (n // len(s)) + s[:n % len(s)].count('a')

  • willy756_taiwan
     + 0 comments

    Python solution

    def repeatedString(s, n):
    a_count = 0
    for c in s:
        if c=="a":
            a_count+=1
    
    repeated = a_count*(n//len(s))
    mod = n%len(s)
    for c in s[:mod]:
        if c=="a":
            repeated+=1
    return repeated