Python program to check for Leap year

In this post, we will learn how to write a program to check for leap year but before writing a program first we should know what is a leap year.

What is Leap year?

A year in which 29 days in the month of February is called a leap year (that means there are 366 days in this year).

We should write this program in three different ways

1. using if-else (In a simple way by taking user input)
2. using function
3. except exception, if the user inputs the wrong data (using try-except)

Now let’s understand how we implement the leap year program with the help of a flowchart.

Flowchart for Leap year

The flowchart helps us to write a program more efficiently and effectively. it is recommended that before writing a program create a flowchart for your program (it helps you a lot).

Algorithm

1. if the year is divisible by 4 then go to step number 2 otherwise year is not a leap year.
2. if the year is divisible by 100 then go to step number 3 otherwise year is leap year.
3. if the year is divisible by 400 then the year is a leap year otherwise it is not a leap year.

Now we understand how to write a program to check for leap year with the help of the above flowchart and algorithm so let’s start writing the program.

1. using if-else (In simple way by taking user input)

Write a program to check for leap year by taking user input.

Before writing this program you should know about:

1. if-else
2. nested if-else

Source code:

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

Output-1:

Enter a year: 2012
2012 is leap year

Output-2:

Enter a year: 2013
2013 is not leap year 

2. Using function

Write A program to check for leap year by using the function.

for understanding this program you should know about:

1. if-else
2. nested if-else
3. function

Source code:

def check_leap_year(year):
    if year % 4 == 0:
        if year % 100 == 0:
            if year % 400 == 0:
                print(f"{year} is a leap year")
            else:
                print(f"{year} is not a leap year")
        else:
            print(f"{year} is a leap year")
    else:
        print(f"{year} is not a leap year")


year = int(input("Enter a year to check for leap year :"))

check_leap_year(year)

Output-1:

Enter a year to check for leap year :2015
2015 is not a leap year

Output-2:

Enter a year to check for leap year :2020
2020 is a leap year

3. except exceptions if user input wrong data

Write a program to check for leap year and also except exception if the user input wrong data with the help of the try-except block.

Before writing this program you should know about:

1. if-else.
2. nested if-else.
3. function.
4. try-except block.

Source code:

def check_leap_year(year):
    if year % 4 == 0:
        if year % 100 == 0:
            if year % 400 == 0:
                print(f"{year} is a leap year")
            else:
                print(f"{year} is not a leap year")
        else:
            print(f"{year} is a leap year")
    else:
        print(f"{year} is not a leap year")


while True:
    try:
        year = int(input("Enter a year to check for leap year :"))
        break
    except Exception:
        print("pls enter numberic value only! and try again")

check_leap_year(year)

Output:

Enter a year to check for leap year :two thousand twenty-seven
pls enter numberic value only! and try again
Enter a year to check for leap year :2027
2027 is not a leap year

Hope this post adds some value to your life -thank you