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any() & len() functions solution in Clear category for Acceptable Password II by tvylormvde
# 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]):
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
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Oct. 5, 2020
