PPA-2

Question

In the Gregorian calendar, a leap year has a total of \(366\) days instead of the usual \(365\) as a result of adding an extra day (February \(29\)) to the year. This calendar was introduced in \(1582\) to replace the flawed Julian calendar. The criteria given below are used to determine if a year is a leap year or not.

• If a year is divisible by \(100\) then it will be a leap year if it is also divisible by \(400\).
• If a year is not divisible by \(100\), then it will be a leap year if it is divisible by \(4\).

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Write a function named check_leap_year that accepts a year between \(1600\) and \(9999\) as argument. It should return True if the year is a leap year and False otherwise.

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You do not have to accept input from the user or print output to the console. You just have to write the function definition.

Hint

Assume that the year is not a leap year to begin with. Now use a binary variable, say leap, that is set to False to begin with. Next, use a bunch of if-else statements to update the value of leap. Finally, return leap. Just to make sure that you understand the conditions, are the following leap years?

• \(1900\)
• \(2024\)
• \(2000\)

Solutions

Solution-1

def check_leap_year(year):
    if year % 100 == 0:
        if year % 400 == 0:
            return True
        return False
    else:
        if year % 4 == 0:
            return True
        return False

Solution-2

def check_leap_year(year):
    if (year % 100 == 0) and (year % 400 == 0):
        return True
    elif (year % 100 != 0) and (year % 4 == 0):
        return True
    return False

Solution-3

def check_leap_year(year):
    if year % 100 == 0:
        return year % 400 == 0
    else:
        return year % 4 == 0

Solution-4

This uses just one return statement at the end of the function.

def check_leap_year(year):
    leap = False
    if year % 100 == 0:
        if year % 400 == 0:
            leap = True
    else:
        if year % 4 == 0:
            leap = True
    return leap

Video Solution