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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Day of the Programmer
  5. Discussions

Day of the Programmer

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  • albiewalbie
     + 19 comments

    Python 2

    def solve(year):
    if (year == 1918):
        return '26.09.1918'
    elif ((year <= 1917) & (year%4 == 0)) or ((year > 1918) & (year%400 == 0 or ((year%4 == 0) & (year%100 != 0)))):
        return '12.09.%s' %year
    else:
        return '13.09.%s' %year

  • axel4u
     + 4 comments

    Hi (java) people. Hope this simple logic will be useful.

    static String solve( int year ) {

    if( year == 1918 ) return "26.09.1918";
    if( isLeap( year ) ) return "12.09." + Integer.toString( year );
    else return "13.09." + Integer.toString( year );

}

static boolean isLeap( int year ) {

    if( year % 4 != 0 ) return false;
    if( year > 1918 && year % 100 == 0 && year % 400 != 0 ) return false;
    return true;

}

  • apghr
     + 1 comment

    C++

    string solve(int y){
    int d = 13;
    if(!(y%4) && (y < 1918 || y%100 || !(y%400))) d--;
    if(y == 1918) d = 26;
    return to_string(d) + ".09." + to_string(y);
}

  • hello_abhi
     + 2 comments

    how is year 1800 a leap year ? acc to the rules mentioned in the problem.

  • ParrotStone
     + 0 comments

    My JS solution

    function dayOfProgrammer(year) {
  if (year == 1918) {
    // Corner case
    return `26.09.${year}`; 
  } else if (year > 1918) {
    // Calculating Leap year in Gregorian Calendar
    const isLeapYear = (year % 4 == 0) && (year % 100 != 0) || year % 400 == 0;
    return `${isLeapYear ? 12 : 13}.09.${year}`;
  } else {
    // Calculating Leap year in Julian Calendar
    return `${year % 4 == 0 ? 12 : 13}.09.${year}`;
  }
}

// dayOfProgrammer(2000) // leap year -> 12...
dayOfProgrammer(1700); // leap year > 13..
// dayOfProgrammer(1800) // NOT leap year -> 12-9-1800
// dayOfProgrammer(2016) // leap year -> 13-9-2016
// dayOfProgrammer(2017) // NOT leap year -> 12-9-2017
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