Check if a Year is a Leap Year or Not in Python

Leap-year-or-not-using-python

Check if a Year is a Leap Year or Not in Python

Given an integer input year, the objective of the code is to Check if a Year is a Leap Year or Not in Python Language. To do so  we’ll check if the year input satisfies either of the two conditions of leap year.

Example
Input : 2020
Output : It's a Leap Year.

Check Whether a Year is a Leap Year or Not in Python

Given an integer input as the year, the objective is to Check if a Year is a Leap Year or Not in Python Language. To do so we’ll check each condition mentioned below in the blue box. It either of the conditions is satisfied, the year is a leap year. It’s not otherwise. Here are some methods to check whether or not it’s a leap year

  • Method 1: Using if-else statements 1
  • Method 2: Using if-else statements 2

We’ll discuss all the above mentioned methods in detail in the upcoming sections. Before going for the approach read out the blue box below for better understanding of the concept.

Method 1: Using if-else Statements 1

In this method we’ll use the if-else statements to check whether or not the input integer satisfies either of the conditions. To learn more about control statements in python, check out if-else control statements in Python.

Working

For a User Input year we do

  1. Check if the year variable is divisible by 400.
  2. Check if the year variable is divisible only by 4 and not 100.
  3. If the above mentioned conditions are satisfied, prints “Leap Year”, print “Not a Leap Year” otherwise.

Let’s implement the above logic in Python Language.

Python Code

Run
year = 2000
if (year%400 == 0) or (year%4==0 and year%100!=0):
  print("Leap Year")
else:
  print("Not a Leap Year")

Output

Leap Year

Method 2: Using if-else Statements 2

In this method we’ll use the if-else statements to check whether or not the input integer satisfies either of the conditions. This method is a modified and simpler version of the previous method.

Working

For a User Input year we do

  1. The input is stored in an int type variable say year.
  2. year is checked for being a leap year or not with the following condition if( ((year % 4 == 0)&&(year % 100 != 0)) || (year % 400==0) )
  3. If the above condition is true then input is a leap year otherwise input is not a leap year.

Let’s implement the above logic in Python Language.

Python Code

Run
year = 2000
if( ((year % 4 == 0) and (year % 100 != 0)) or (year % 400==0) ):
    print("Leap Year")
else:
    print("Not leap Year")

Output:

Leap Year

35 comments on “Check if a Year is a Leap Year or Not in Python”


  • M

    year = int(input(“enter the year:”))
    if (year % 4 == 0 and year % 400 != 0)or(year % 400 == 0):
    print(“yes {} is leap year”.format(year))
    else:
    print(“no {} is not a leap year”.format(year))


  • murali

    import calendar
    year = int(input(“enter the year”))
    if (calendar.isleap(year)):
    print(“{} is leap year”.format(year))
    else:
    print(“{} not a leap year”.format(year))


  • bhaskar

    year=int(input(“Enter year”))
    if(year%4==0 and year!=100) or year%400==0:
    print(“leap year”)
    else:
    print(“not a leap year”)


  • 18131a0437

    year=int(input())
    if year%4==0:
    if year%400==0 or year%100!=0:
    print(year,’is Leap Year’)
    else:
    print(year,’is not a Leap Year’)
    else:
    print(year,’is not a Leap Year’)


  • Neha

    y=int(input())
    if(y%4==0 and y%100==0 or y%400==0):
    print(“yes {} is a leap year”.format(y))

    else:
    print(“no {} is not a leap year”.format(y))


  • Mancy

    Can be made more simple:
    year = int(input(“Enter the year: “))
    if year % 4 == 0 and year % 100 != 0 or year % 400 == 0:
    print(“{} is leap year”.format(year))
    else:
    print(“{} is not leap year”.format(year))


  • rohith

    year=int(input(“enter year”))
    if(year%4==0 and year%100!=0 or year%400==0):
    print(“leap year”)
    else:
    print(“not leap year”)


  • Samiksha

    year = int(input(“enter the year: “))

    if year % 4 == 0:
    if year % 100 != 0:
    print(“Leap Year”)
    else:
    if year % 400 != 0:
    print(“Leap Year”)

    else:
    print(“Not a Leap Year”)


  • gaurav

    more shorter:

    year = int(input(“enter year: “))
    if (year % 4) == 0:
    if (year % 400):
    print(“non-leap year”)
    exit()
    print(“leap year”)
    else:
    print(“Non-leap year”)


  • premkumar

    def leapyear(year):
    import calendar
    return (calendar.isleap(year))

    year = int(input())
    if (leapyear(year)):
    print(“Leap Year”)
    else:
    print(“Not a Leap Year”)


  • pavan

    easy one:
    Bhavantik Correct one:
    year = int(input(“Enter a Year: “))
    if year%4 == 0:
    if year%100==0 and year%400 !=0:
    print(year,” is not a Leap Year”)
    else:
    print(year,” is Leap Year”)
    else:
    print(year,” is not a Leap Year”)


    • pavan

      year=int(input(“Enter the year”))
      if(year%4==0):
      print(“Leap year”)
      elif(year%4==0 and year%100!=0):
      print(“leap year”)
      elif(year%100==0 and year%400!=0):
      print(“LEap year”)
      else:
      print(“not a leap year”)


  • Yashi

    year = int(input(“Enter Year:”))
    if year % 4 == 0:
    if year % 100 == 0 and year%400 !=0:
    print(“No, {} is not a Leap Year”.format(year))
    else:
    print(“Yes, {} is Leap Year”.format(year))
    else:
    print(“No, {} is not a Leap Year”.format(year))