Check if a Year is a Leap Year

Determining whether a year is a leap year is a common programming task, especially when working with dates and calendars. Python offers straightforward ways to accomplish this, and this post will look at many approaches, from basic conditional logic to using the calendar module.

Understanding Leap Years

Before diving into the code, let’s briefly review the rules for leap years:

• A leap year is divisible by 4.
• However, if a year is divisible by 100, it’s NOT a leap year, unless…
• It’s also divisible by 400.

This seemingly complex rule can be elegantly translated into Python code.

Method 1: Using Conditional Logic

This method directly implements the leap year rules using if-elif-else statements. It’s clear, concise, and easy to understand.

def is_leap_year(year):
  """Checks if a year is a leap year using conditional logic."""
  if year % 4 == 0:
    if year % 100 == 0:
      if year % 400 == 0:
        return True
      else:
        return False
    else:
      return True
  else:
    return False

# Example usage
print(is_leap_year(2024))  # Output: True
print(is_leap_year(2000))  # Output: True
print(is_leap_year(1900))  # Output: False
print(is_leap_year(2023))  # Output: False

Method 2: A More Concise Approach with Conditional Expressions

We can improve readability and reduce code length by using Python’s conditional expressions (ternary operator).

def is_leap_year_concise(year):
  """Checks if a year is a leap year using conditional expressions."""
  return year % 4 == 0 and (year % 100 != 0 or year % 400 == 0)

# Example usage
print(is_leap_year_concise(2024))  # Output: True
print(is_leap_year_concise(2000))  # Output: True
print(is_leap_year_concise(1900))  # Output: False
print(is_leap_year_concise(2023))  # Output: False

This version cleverly uses the and and or operators to concisely express the leap year rules.

Method 3: Leveraging the calendar Module

Python’s built-in calendar module provides a convenient function for this purpose.

import calendar

def is_leap_year_calendar(year):
    """Checks if a year is a leap year using the calendar module."""
    return calendar.isleap(year)

# Example usage
print(is_leap_year_calendar(2024))  # Output: True
print(is_leap_year_calendar(2000))  # Output: True
print(is_leap_year_calendar(1900))  # Output: False
print(is_leap_year_calendar(2023))  # Output: False

The calendar.isleap() function provides a simple and efficient way to check for leap years. This is often the preferred method due to its clarity and reliance on a well-tested library function.

Handling Invalid Input

For code clarity, consider adding error handling to manage invalid input, such as non-integer values or negative years. This could involve using a try-except block to catch TypeError or ValueError exceptions. We leave this as an exercise for the reader.