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final solution: finds total days between given dates
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NirmalSilwal committed Apr 26, 2020
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# Use Dave's suggestions to finish your daysBetweenDates
# procedure. It will need to take into account leap years
# in addition to the correct number of days in each month.


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


def daysInMonth(year, month):
if month == 1 or month == 3 or month == 5 or month == 7 or month == 8 or month == 10 or month == 12:
return 31
else:
if month == 2:
if leapYear(year):
return 29
else:
return 28
else:
return 30


def nextDay(year, month, day):
"""Simple version: assume every month has 30 days"""
if day < daysInMonth(year, month):
return year, month, day + 1
else:
if month == 12:
return year + 1, 1, 1
else:
return year, month + 1, 1


def dateIsBefore(year1, month1, day1, year2, month2, day2):
"""Returns True if year1-month1-day1 is before year2-month2-day2. Otherwise, returns False."""
if year1 < year2:
return True
if year1 == year2:
if month1 < month2:
return True
if month1 == month2:
return day1 < day2
return False


def daysBetweenDates(year1, month1, day1, year2, month2, day2):
"""Returns the number of days between year1/month1/day1
and year2/month2/day2. Assumes inputs are valid dates
in Gregorian calendar."""
# program defensively! Add an assertion if the input is not valid!

assert not dateIsBefore(year2, month2, day2, year1, month1, day1)

days = 0

while dateIsBefore(year1, month1, day1, year2, month2, day2):
year1, month1, day1 = nextDay(year1, month1, day1)
days += 1
return days


def test():
test_cases = [((2012, 1, 1, 2012, 2, 28), 58),
((2012, 1, 1, 2012, 3, 1), 60),
((2011, 6, 30, 2012, 6, 30), 366),
((2011, 1, 1, 2012, 8, 8), 585),
((1900, 1, 1, 1999, 12, 31), 36523)]

for (args, answer) in test_cases:
result = daysBetweenDates(*args)
if result != answer:
print("Test with data:", args, "failed")
else:
print("Test case passed!")


test()

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