Days in Month

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Solution

Days in Month: Solution

Given a string with a month and a year (separated by a space), return the number of days in that month.

Leap years are a bit tricky. A year is a leap year if and only if:

• it is evenly divisible by 4

• except if it is divisible by 100, in which case it isn’t

• except if it is divisible by 400, in which case it is

So, for example, 1904 was a leap year. 1900 is divisible by 100, so it wasn’t. 2000 is divisible by 400, so it was.

To help with this, we’ve given you a function to determine if a year is a leap year; you may use this in your solution:

def is_leap_year(year):
    """Is this year a leap year?

    Every 4 years is a leap year::

        >>> is_leap_year(1904)
        True

    Except every hundred years::

        >>> is_leap_year(1900)
        False

    Except-except every 400::

        >>> is_leap_year(2000)
        True
    """

    if year % 400 == 0:
        return True

    if year % 100 == 0:
        return False

    if year % 4 == 0:
        return True

>>> for i in range(1, 13):
...     date = str(i) + " 2016"
...     print "%s has %s days." % (date, days_in_month(date))
1 2016 has 31 days.
2 2016 has 29 days.
3 2016 has 31 days.
4 2016 has 30 days.
5 2016 has 31 days.
6 2016 has 30 days.
7 2016 has 31 days.
8 2016 has 31 days.
9 2016 has 30 days.
10 2016 has 31 days.
11 2016 has 30 days.
12 2016 has 31 days.

>>> days_in_month("02 2015")
28

def days_in_month(date):
    """How many days are there in a month?"""