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~~Counting Sundays without Python date module~~

Counting Sundays without Python datetime module

# The problem
> You are given the following information, but you may prefer to do some
> research for yourself.
>
> * 1 Jan 1900 was a Monday.
> * Thirty days has September, April, June and November. All the rest have thirty-one, saving February alone, which has twenty-eight, rain
> or shine.
> * And on leap years, twenty-nine. A leap year occurs on any year evenly divisible by 4, but not on a century unless it is divisible by 400.
>
> How many Sundays fell on the first of the month during the twentieth
> century (1 Jan 1901 to 31 Dec 2000)?
# The solution
```
def is_century(year):
return year % 100 == 0
def is_leap_year(year):
if is_century(year):
return year % 400 == 0
else:
return year % 4 == 0
def get_day_count(year, month):
if month == 2 and is_leap_year(year):
return 29
elif month == 2:
return 28
elif month in [4, 6, 9, 11]:
return 30
else:
return 31
def solution(start_day, start_month, start_year, end_day, end_month, end_year):
current_day, current_month, current_year = start_day, start_month, start_year
days = 0
sundays = 0
while current_day != end_day or current_month != end_month or current_year != end_year:
if days % 7 == 0 and current_day == 1:
sundays += 1
days += 1
if current_day == get_day_count(current_year, current_month):
current_day = 1
if current_month == 12:
current_month = 1
current_year += 1
else:
current_month += 1
else:
current_day += 1
return sundays
print(solution(1, 1, 1901, 31, 12, 2000))
```
~~I solved Project Euler Problem 19 with Python. My goal was to solve this without the `date` module, and handling the calendar part on my own.~~
As the problem states the start date is a Monday, I counted the next Mondays that fell on the first of the month using this condition: `if days % 7 == 0 and current_day == 1:`, being `days` the number of days I counted so far inside the loop.
What can be improved? Have I overcomplicated some part of the code?

# The problem
> You are given the following information, but you may prefer to do some
> research for yourself.
>
> * 1 Jan 1900 was a Monday.
> * Thirty days has September, April, June and November. All the rest have thirty-one, saving February alone, which has twenty-eight, rain
> or shine.
> * And on leap years, twenty-nine. A leap year occurs on any year evenly divisible by 4, but not on a century unless it is divisible by 400.
>
> How many Sundays fell on the first of the month during the twentieth
> century (1 Jan 1901 to 31 Dec 2000)?
# The solution
```
def is_century(year):
return year % 100 == 0
def is_leap_year(year):
if is_century(year):
return year % 400 == 0
else:
return year % 4 == 0
def get_day_count(year, month):
if month == 2 and is_leap_year(year):
return 29
elif month == 2:
return 28
elif month in [4, 6, 9, 11]:
return 30
else:
return 31
def solution(start_day, start_month, start_year, end_day, end_month, end_year):
current_day, current_month, current_year = start_day, start_month, start_year
days = 0
sundays = 0
while current_day != end_day or current_month != end_month or current_year != end_year:
if days % 7 == 0 and current_day == 1:
sundays += 1
days += 1
if current_day == get_day_count(current_year, current_month):
current_day = 1
if current_month == 12:
current_month = 1
current_year += 1
else:
current_month += 1
else:
current_day += 1
return sundays
print(solution(1, 1, 1901, 31, 12, 2000))
```
I solved Project Euler Problem 19 with Python. My goal was to solve this without the `datetime` module, and handling the calendar part on my own.
As the problem states the start date is a Monday, I counted the next Mondays that fell on the first of the month using this condition: `if days % 7 == 0 and current_day == 1:`, being `days` the number of days I counted so far inside the loop.
What can be improved? Have I overcomplicated some part of the code?

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