Python - Find if given year is Leap Year in Georgian Calendar

Python - Check if year is Leap Year in Georgian Calendar

In this tutorial, we will write a Python function to check if the given year is a leap year in Georgian Calendar.

Following are the criteria for an year to be Leap year.

The year can be evenly divided by 4, is a leap year, unless:
- The year can be evenly divided by 100, it is NOT a leap year, unless:
  - The year is also evenly divisible by 400. Then it is a leap year.

You may combine all these conditions into a single boolean expression.

Consider that we have our year in variable year.

To satisfy the first condition, "The year can be evenly divided by 4, is a leap year", following condition has to be satisfied.

year%4==0

To satisfy the second condition, "The year can be evenly divided by 100, it is NOT a leap year,", following condition has to be satisfied.

year%100==0

To satisfy the third condition, " The year is also evenly divisible by 400", following condition has to be satisfied.

year%400==0

By combining above three conditions, we get the following boolean expression.

year%4==0 and not (year%100==0 and not year%400==0)

unless in the given condition translates to and not and hence the and not operators in the above condition.

Following is the example program, in which we check if a given year is a leap year or not.

Python Program

def is_leap(year):
    return year%4==0 and not (year%100==0 and not year%400==0)

year = int(input())

print(is_leap(year))

Output

2004
True

When you run the program, input() waits for your input. Enter a number and our program shall check if the given number is a leap year or not.

In this tutorial, we learned how to check if given year is a leap year of Georgian Calendar or not.