An extra day is added to the calendar almost every four years as February 29, and the day is called a *leap day*. It corrects the calendar for the fact that our planet takes approximately 365.25 days to orbit the sun. A leap year contains a leap day.

In the Gregorian calendar, three conditions are used to identify leap years:

- 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.

- The year can be evenly divided by 100, it is NOT a leap year, unless:

This means that in the Gregorian calendar, the years 2000 and 2400 are leap years, while 1800, 1900, 2100, 2200, 2300 and 2500 are NOT leap years. Source

**Task**

Given a year, determine whether it is a leap year. If it is a leap year, return the Boolean `True`

, otherwise return `False`

.

Note that the code stub provided reads from STDIN and passes arguments to the `is_leap`

function. It is only necessary to complete the `is_leap`

function.

**Input Format**

Read **year, **the year to test.

**Constraints**

1900 <= y <=10^5

**Output Format**

The function must return a Boolean value (True/False). Output is handled by the provided code stub.

**Sample Input 0**

1990

**Sample Output 0**

False

**Explanation 0**

1990 is not a multiple of 4 hence it’s not a leap year.

### Write a function in Python – Hacker Rank Solution

def is_leap(year): leap = False # Write your logic here # Write a function in Python - Hacker Rank Solution START if year%400==0 : leap = True; elif year%4 == 0 and year%100 != 0: leap = True; return leap # Write a function in Python - Hacker Rank Solution END year = int(raw_input()) print is_leap(year)

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