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one liner using any() and all() solution in Clear category for Acceptable Password III by pacurar.sebastian90
def is_acceptable_password(password: str) -> bool:
'''
Verify True if length is bigger than 6, there is at least one digit in string using any(),
and not all string characters are digits using all(), else False
'''
return True if len(password) > 6 and any([i.isdigit() for i in password]) and not all([i.isdigit() for i in password]) else 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
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Feb. 28, 2021
