HackerRank Python: Write a function Solution

Table of Contents

Question

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.

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≤year≤10^5

Output Format

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

Sample Input 0

Sample Output 0

False

Explanation 0

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

Solution

				
					def is_leap(year):
    leap = False   
    if year%100==0 and year%400==0:
        leap = True
    elif year%4==0 and year%100!=0:
        leap = True
    else:
        leap = False
    
    return leap

year = int(input())
print(is_leap(year))

Explanation

In this code, we define the is_leap function that takes the year as input and checks the three conditions to determine whether it’s a leap year or not. We use the modulo operator (%) to check for divisibility.

Note: The code assumes that the input year is within the given constraints (1900 ≤ year ≤ 10^5).

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