# Check if a Year is a Leap Year or Not in 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))

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

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))