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Calculate two year's worth of birthdays solution in Clear category for Next Birthday by kkkkk
from typing import Dict, Tuple
from calendar import isleap
from datetime import date
Date = Tuple[int, int, int]
def next_birthday(today: Date, birthdates: Dict[str, Date]) -> \
Tuple[int, Dict[str, int]]:
"""For upcoming closest bday, return days to bday and persons with bday."""
todays_date = date(*today)
todays_year = todays_date.year
# Calculate birthdays coming up this year and for the next. Depending
# on "today"s date, some birthdays may fall in the next year.
bdays = {todays_year: {}, todays_year+1: {}}
for person, birthdate in birthdates.items():
for year in bdays:
# If the birthdate is 2/29 and "today" is not in a leap year,
# switch, the birthdate to 3/1.
_, month, day = birthdate
if month == 2 and day == 29 and not isleap(year):
month, day = 3, 1
# Calculate the number of days between the birthdate and the
# given date of today. Only save upcoming birthdays.
days_till_bday = date(year, month, day) - todays_date
if days_till_bday.days >= 0:
bdays[year][person] = days_till_bday.days
# If there are birthdays for "today"s year, that's the year we'll
# use because they'll come sooner than next year's.
if not bdays[todays_date.year]:
todays_year = todays_date.year + 1
# Find the closest birthday, then all the people with that birthday.
closest_day = min(bdays[todays_year].values())
bday_persons = {}
for person, days in bdays[todays_year].items():
if days == closest_day:
bday_persons[person] = (todays_year - birthdates[person][0])
return (closest_day, bday_persons)
Sept. 15, 2020
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