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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Library Fine
  5. Discussions

Library Fine

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1032 Discussions

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  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can watch the explanation here : https://youtu.be/EhPgZkwwx4Y

    int libraryFine(int d1, int m1, int y1, int d2, int m2, int y2) {
    int res = 0;
    if(y1<y2 || (y1 == y2 && m1<m2) || (y1 == y2 && m2 == m1 && d1<=d2)) return 0;
    if(y2 == y1){
        if(m1 == m2) return 15 * (d1 - d2);
        else return (m1 - m2) * 500;
    }
    else return 10000;
    return 0;
}

  • akash_prajapati6
     + 0 comments
        if(y1==y2 and m1==m2 and d1>d2):
        return 15*(d1-d2)
    elif(y1==y2 and m1>m2):
        return 500*(m1-m2)
    elif(y1>y2):
        return 10000
    else:
        return 0

  • abhijeetmaharan2
     + 0 comments
    def libraryFine(d1, m1, y1, d2, m2, y2):
    # If the book is returned after the expected year
    if y1 > y2:
        return 10000
    # If the book is returned in the same year but after the expected month
    elif y1 == y2 and m1 > m2:
        return (m1 - m2) * 500
    # If the book is returned in the same year and month but after the expected day
    elif y1 == y2 and m1 == m2 and d1 > d2:
        return (d1 - d2) * 15
    # If the book is returned on or before the due date
    else:
        return 0

  • jjr20060326
     + 0 comments
    def libraryFine(d1, m1, y1, d2, m2, y2):
    if y1 > y2:
        return 10000
    if y1 == y2:
        if m1 > m2:
            return 500 * (m1 - m2)
        if m1 == m2 and d1 > d2:
            return 15 * (d1 - d2)
    return 0

  • patrickgao2020
     + 0 comments

    Here is my Python solution! It looks a bit complicated but is just a series of if, elif, and else statements to calculate the fine.

    def libraryFine(d1, m1, y1, d2, m2, y2):
    if y1 < y2:
        return 0
    elif y1 == y2:
        if m1 < m2:
            return 0
        elif m1 == m2:
            if d1 <= d2:
                return 0
            else:
                return 15 * (d1 - d2)
        else:
            return 500 * (m1 - m2)
    else:
        return 10000