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using any(), all() solution in Clear category for Acceptable Password III by tvylormvde
# Taken from mission Acceptable Password II
# Taken from mission Acceptable Password I
def is_acceptable_password(password: str) -> bool:
if len(password) > 6 and any([i.isdigit() for i in password]):
if not all([i.isdigit() for i in password]):
return True
return False
if __name__ == '__main__':
print("Example:")
print(is_acceptable_password('short'))
# These "asserts" are used for self-checking and not for an auto-testing
assert is_acceptable_password('short') == False
assert is_acceptable_password('muchlonger') == False
assert is_acceptable_password('ashort') == False
assert is_acceptable_password('muchlonger5') == True
assert is_acceptable_password('sh5') == False
assert is_acceptable_password('1234567') == False
print("Coding complete? Click 'Check' to earn cool rewards!")
Oct. 5, 2020
