Python Program to Check Whether a given Year is a Leap Year or not (3 different ways)

In this article, we will discuss how to write a python program to check whether a given year is a leap year or not.

Firstly, let's discuss "What is Leap year?".

A year is a Leap year if it is exactly divisible by 4 except for the years ending with double zero(00) (also called century years). And among the century years, only those years are the leap year which is perfectly divisible by 400.

I will be discussing 3 different ways to write the python program for it.

  1. Using multiple if-else statements
  2. Using single if-else statement
  3. And by using if-elif statement

Let's discuss all the ways one by one.


Python Program to Check Whether a given Year is a Leap Year or not

Using multiple if-else statements

In this method, we will be using multiple if-else statements.

I have also used format() function so that I would be able to print the output in a formatted way.

In this program, the {0} is replaced with the variable passed inside the parameters of format() function. In this example, the 0 inside the braces({}) represents the parameter number to be replaced with.

Code:-

# taking year in YYYY format
year = int(input("Enter the year: "))                           

# checking whether year is divisible by 4
if(year%4 == 0):                                                
    # checking whether year is divisible by 100
    if(year%100 == 0):                                          
        # checking whether year is divisible by 400
        if(year%400 == 0):                                      
            # printing the year is leap year
            print("{0} is a leap year.".format(year))       
        else:
            # printing the year is not a leap year
            print("{0} is not a leap year.".format(year))   
    else:
        # printing the year is leap year
        print("{0} is a leap year.".format(year))           
else:
    # printing the year is not a leap year
    print("{0} is not a leap year.".format(year))          

Output:-

Enter the year: 2016
2016 is a leap year.



Using single if-else statement

We can shorten the above program by using just single if-else statement.

Here also I have used the format() function.

Code:-

# taking year in YYYY format
year = int(input("Enter the year: "))                           

# checking whether year is leap year or not
if year%400 == 0 or year%4 == 0 and year%100 != 0:
    print("{0} is a leap year.".format(year))
else:
    print("{0} is not a leap year.".format(year))

Output:-

Enter the year: 2100
2100 is not a leap year.



Using if-elif statement

Here, we will be discussing the use the if-elif statement. The elif is short for else-if.

Code:-

# taking year in YYYY format
year = int(input("Enter the year: "))                           

# checking whether year is divisible by 400  
if(year%400 == 0):                                      
    # printing the year is leap year
    print("{0} is a leap year.".format(year))  
# checking whether year is divisible by 100     
elif(year%100 == 0):                                          
    # printing the year is not a leap year
    print("{0} is not a leap year.".format(year))   
# checking whether year is divisible by 4
elif(year%4 == 0):
    # printing the year is leap year
     print("{0} is a leap year.".format(year))           
else:
    # printing the year is not a leap year
    print("{0} is not a leap year.".format(year))

Output:-

Enter the year: 1300
1300 is not a leap year.

Hence, we have discussed all the ways.

We have learned different ways to use the if-else statements.

And also we have learned about the use of format() function.

You can learn about many other Python Programs Here.







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Amit Rawat

Amit Rawat

Founder and Developer at SpiderLabWeb
I love to work on new projects and also work on my ideas. My main field of interest is Web Development.

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