Python Program to Check if Year is Leap Year

You can check if a year is a leap year in Python using if (year % 400 == 0) or (year % 4 == 0 and year % 100 != 0): to test the leap year conditions.

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 decide if a year is leap, check if it divides evenly by 400, or if it divides evenly by 4 but not by 100. This matches the real-world leap year rules.

Algorithm

1

Get the year as input

2

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

3

Otherwise, check if the year is divisible by 4 but not by 100; if yes, it is a leap year

4

If none of the above, 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 divisible by 400

1900 % 400 == 300 (False)

3

Check divisible 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 year

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 by 100 means leap year

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

Step 3: Otherwise, not a leap year

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 a function

python

def is_leap_year(year):
    return (year % 400 == 0) or (year % 4 == 0 and year % 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 wraps the logic in a function for reuse and clarity.

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

Time Complexity

The program performs a fixed number of arithmetic and logical operations regardless of input size, so it runs in constant time O(1).

Space Complexity

Only a few variables are used, and no extra data structures are created, so space complexity is O(1).

Which Approach is Fastest?

All approaches run in constant time, but using the calendar module adds a small overhead due to import, while the direct arithmetic check is fastest and simplest.

Approach	Time	Space	Best For
Direct arithmetic check	O(1)	O(1)	Simple and fast checks without imports
calendar.isleap() function	O(1)	O(1)	Using built-in library for readability
Function wrapper	O(1)	O(1)	Reusable code and clarity