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datetime timedelta strftime strptime solution in Clear category for The First Working Day by kdim
from datetime import datetime, timedelta
def vacation(dt, dy):
d = datetime.strptime(dt, '%Y-%m-%d') + timedelta(days=dy)
d += timedelta(days=(7 - d.weekday()) * (d.weekday() > 4))
return d.strftime('%Y-%m-%d')
if __name__ == '__main__':
print("Example:")
print(vacation('2018-07-01', 14))
assert vacation('2018-07-01', 14) == '2018-07-16'
assert vacation('2018-02-19', 10) == '2018-03-01'
assert vacation('2000-02-28', 5) == '2000-03-06'
assert vacation('1999-12-20', 14) == '2000-01-03'
print("Coding complete? Click 'Check' to earn cool rewards!")
Feb. 25, 2021
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