L10Q21_DefineSimpleNextDay.py

###
### Define a simple nextDay procedure, that assumes
### every month has 30 days.
###
### For example:
###    nextDay(1999, 12, 30) => (2000, 1, 1)
###    nextDay(2013, 1, 30) => (2013, 2, 1)
###    nextDay(2012, 12, 30) => (2013, 1, 1)  (even though December really has 31 days)
###

def nextDay(year, month, day):
    """
    Returns the year, month, day of the next day.
    Simple version: assume every month has 30 days.
    """
    '''
    #given solution is more elegant
    if day < 30:
        return year, month, day + 1
    else:
        if month < 12:
            return year, month + 1, 1
        else:
            return year + 1, 1, 1
'''
#my initial solution, it works but is longer than provided

    if day < 30:
        day += 1
    if day == 30 or day > 30:
        day = 1
        if month < 12:
            month += 1
        if month == 12:
            month = 1
            year += 1
    return (year,month,day)

#test_cases
print nextDay(1999, 12, 30) # (2000, 1, 1)
print nextDay(2013, 1, 30) # (2013, 2, 1)
print nextDay(2012, 12, 30) #(2013, 1, 1)(even though December really has 31 days)
print nextDay(2012, 12, 31)
''' more tests
nextDay(2012, 4, 30)
nextDay(2012, 12, 1)
nextDay(1999, 12, 30)
nextDay(2012, 12, 30)'''