How to find the year is leap year or not in python

View Discussion

Improve Article

Save Article

View Discussion

Improve Article

Save Article

A year is a leap year if the following conditions are satisfied: 

1. The year is multiple of 400.
2. The year is multiple of 4 and not multiple of 100.

Leap year or not.

pseudo-code 

• if year is divisible by 400 then is_leap_year
• else if year is divisible by 100 then not_leap_year
• else if year is divisible by 4 then is_leap_year
• else not_leap_year

Example 1: Checking all the possible conditions for a leap year.

C++

#include

using namespace std;

bool checkYear(int year)

{

    if (year % 400 == 0)

        return true;

    if (year % 100 == 0)

        return false;

    if (year % 4 == 0)

        return true;

    return false;

}

int main()

{

    int year = 2000;

    checkYear(year) ? cout << "Leap Year":

                      cout << "Not a Leap Year";

    return 0;

}

C

#include

#include

bool checkYear(int year)

{

    if (year % 400 == 0)

        return true;

    if (year % 100 == 0)

        return false;

    if (year % 4 == 0)

        return true;

    return false;

}

int main()

{

    int year = 2000;

    checkYear(year)? printf("Leap Year"):

                   printf("Not a Leap Year");

    return 0;

}

Java

class Test

{

    static boolean checkYear(int year)

    {

        if (year % 400 == 0)

            return true;

        if (year % 100 == 0)

            return false;

        if (year % 4 == 0)

            return true;

        return false;

    }

    public static void main(String[] args)

    {

        int year = 2000;

        System.out.println( checkYear(2000)? "Leap Year" :

                           "Not a Leap Year" );

    }

}

Python3

def checkYear(year):

    if (year % 4) == 0:

        if (year % 100) == 0:

            if (year % 400) == 0:

                return True

            else:

                return False

        else:

             return True

    else:

        return False

year = 2000

if(checkYear(year)):

    print("Leap Year")

else:

    print("Not a Leap Year")

C#

using System;

class GFG

{

    static bool checkYear(int year)

    {

        if (year % 400 == 0)

            return true;

        if (year % 100 == 0)

            return false;

        if (year % 4 == 0)

            return true;

        return false;

    }

    public static void Main()

    {

        int year = 2000;

        Console.Write( checkYear(year)? "Leap Year" :

                                 "Not a Leap Year" );

    }

}

PHP

function checkYear($year)

{

    if ($year % 400 == 0)

        print("Leap Year");

    else if ($year % 100 == 0)

        print("Not a Leap Year");

    else if ($year % 4 == 0)

        print("Leap Year");

    else

        print("Not a Leap Year");

}

$year = 2000;

checkYear($year);

?>

Javascript

Output:

Leap Year

Time Complexity : O(1)

Auxiliary Space: O(1)

Example 2: One line code for checking whether the year is a leap year or not.

C++

#include

using namespace std;

bool checkYear(int year)

{

    return (((year % 4 == 0) && (year % 100 != 0)) ||

             (year % 400 == 0));

}

int main()

{

    int year = 2000;

    checkYear(year)? cout << "Leap Year":

                     cout << "Not a Leap Year";

    return 0;

}

C

#include

#include

bool checkYear(int year)

{

return (((year % 4 == 0) && (year % 100 != 0)) ||

        (year % 400 == 0));

}

int main()

{

    int year = 2000;

    checkYear(year)? printf("Leap Year"):

                printf("Not a Leap Year");

    return 0;

}

Java

class Test

{

    static boolean checkYear(int year)

    {

    return (((year % 4 == 0) && (year % 100 != 0)) ||

            (year % 400 == 0));

    }

    public static void main(String[] args)

    {

        int year = 2000;

        System.out.println(checkYear(2000)? "Leap Year" :

                           "Not a Leap Year" );

    }

}

Python3

def checkYear(year):

    return (((year % 4 == 0) and (year % 100 != 0)) or (year % 400 == 0));

year = 2000

if(checkYear(year)):

    print("Leap Year")

else:

    print("Not a Leap Year")

C#

using System;

class GFG

{

    static bool checkYear(int year)

    {

        return (((year % 4 == 0) && (year % 100 != 0)) ||

                (year % 400 == 0));

    }

    public static void Main()

    {

        int year = 2000;

        Console.Write( checkYear(year)? "Leap Year" :

                                 "Not a Leap Year" );

    }

}

PHP

function checkYear($year)

{

    return ((($year % 4 == 0) &&

             ($year % 100 != 0)) ||

             ($year % 400 == 0));

}

$year = 2000;

checkYear($year)? print("Leap Year"):

                  print("Not a Leap Year");

?>

Javascript

Output:

Leap Year

Time Complexity: O(1)

Auxiliary Space: O(1)

Example 3: Check Leap Year using Macros in C/C++

C++

#include

using namespace std;

#define ISLP(y) ((y % 400 == 0) ||\

(y % 100 != 0) && (y % 4 == 0))

int main()

{

    int year = 2020;

    cout << ISLP(year) << "\n";

    return 0;

}

Java

public class Main

{

  static int ISLP(int y)

  {

    if((y % 400 == 0) ||

       (y % 100 != 0) &&

       (y % 4 == 0))

    {

      return 1;

    }

    else

    {

      return 0;

    }

  }

  public static void main(String[] args)

  {

    int year = 2020;

    System.out.println(ISLP(year));

  }

}

Python3

def ISLP(y):

  if((y % 400 == 0) or

     (y % 100 != 0) and

     (y % 4 == 0)):

    return 1;

  else:

    return 0;

if __name__=='__main__':

  year = 2020;

  print(ISLP(year));

C#

using System;

class GFG {

  static int ISLP(int y)

  {

    if((y % 400 == 0) ||

       (y % 100 != 0) &&

       (y % 4 == 0))

    {

      return 1;

    }

    else

    {

      return 0;

    }

  }

  static void Main()

  {

    int year = 2020;

    Console.WriteLine(ISLP(year));

  }

}

Javascript

Output:

1

Time Complexity : O(1)

Auxiliary Space: O(1)

Explanation:

The program outputs 1 if the year is a leap and 0 if it’s not a leap year.

Example 4: Short Solution in Python 

Python

def checkYear(year):

    import calendar

    return(calendar.isleap(year))

year = 2000

if (checkYear(year)):

    print("Leap Year")

else:

    print("Not a Leap Year")

Output:

Leap Year

Time Complexity : O(1)

Auxiliary Space: O(1)

Example 5: Short Solution in Java using isLeap() method of Year class

Java

import java.io.*;

import java.time.Year;

class GFG {

    public static boolean checkYear(int year){

        Year y = Year.of(year);

        return y.isLeap(); 

    }

    public static void main (String[] args) {

        int year = 2000;

        if(checkYear(year)){ 

            System.out.println("Leap Year");

        }else{

            System.out.println("Not a Leap Year");

        }

    }

}

Output:

Leap Year

Time Complexity : O(1)

Auxiliary Space: O(1)

Explanation:

Year class in java is an in-built class that is used to represent an year as a date-time immutable object. The idea is to invoke the isLeap() method of the class which returns true if the year is a leap year and vice-versa.

Below is the implementation of the above idea: