Python Program to Check Leap Year

You can check a leap year in Python by using the condition year % 400 == 0 or (year % 4 == 0 and year % 100 != 0) which returns True if the year is leap, otherwise False.

Examples

Input2000

Output2000 is a leap year

Input1900

Output1900 is not a leap year

Input2024

Output2024 is a leap year

How to Think About It

To check if a year is leap, first see if it is divisible by 400. If yes, it is leap. If not, check if it is divisible by 4 but not by 100. If yes, it is leap. Otherwise, it is not leap.

Algorithm

1

Get the year as input

2

Check if the year is divisible by 400; if yes, it is leap

3

Else check if the year is divisible by 4 and not divisible by 100; if yes, it is leap

4

Otherwise, it is not a leap year

5

Print the result

Code

python

year = int(input('Enter a year: '))
if year % 400 == 0 or (year % 4 == 0 and year % 100 != 0):
    print(f'{year} is a leap year')
else:
    print(f'{year} is not a leap year')

Output

Enter a year: 2024
2024 is a leap year

Dry Run

Let's trace the year 1900 through the code

1

Input year

year = 1900

2

Check divisibility by 400

1900 % 400 == 300 (False)

3

Check divisibility by 4 and not by 100

1900 % 4 == 0 (True) and 1900 % 100 == 0 (True), so condition is False

4

Result

Print '1900 is not a leap year'

Step	Condition	Result
Divisible by 400?	1900 % 400 == 0	False
Divisible by 4 and not 100?	1900 % 4 == 0 and 1900 % 100 != 0	False

Why This Works

Step 1: Divisible by 400 means leap

If a year divides evenly by 400, it is always a leap year, so we check year % 400 == 0 first.

Step 2: Divisible by 4 but not 100 means leap

If the year is divisible by 4 but not by 100, it is a leap year, so we check year % 4 == 0 and year % 100 != 0.

Step 3: Otherwise not leap

If neither condition is true, the year is not a leap year.

Alternative Approaches

Using calendar module

python

import calendar
year = int(input('Enter a year: '))
if calendar.isleap(year):
    print(f'{year} is a leap year')
else:
    print(f'{year} is not a leap year')

This uses Python's built-in calendar module for simplicity but adds an import dependency.

Using function with return

python

def is_leap_year(y):
    return y % 400 == 0 or (y % 4 == 0 and y % 100 != 0)
year = int(input('Enter a year: '))
if is_leap_year(year):
    print(f'{year} is a leap year')
else:
    print(f'{year} is not a leap year')

This approach separates logic into a function for reuse and clarity.

Complexity: O(1) time, O(1) space

Time Complexity

The program performs a fixed number of arithmetic operations and comparisons, so it runs in constant time.

Space Complexity

Only a few variables are used, so the space used is constant.

Which Approach is Fastest?

All approaches run in constant time; using the calendar module adds slight overhead but improves readability.

Approach	Time	Space	Best For
Direct condition check	O(1)	O(1)	Simple and fast checks
Using calendar module	O(1)	O(1)	Readability and built-in reliability
Function with return	O(1)	O(1)	Code reuse and clarity