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First solution in Clear category for Acceptable Password V by MouseCool123
def is_acceptable_password(Password: str) -> bool:
return len(Password) > 6 and (0 < len([Element for Element in Password if Element.isdigit()]) < len(Password) or len(Password) > 9) and Password.lower().find("password") == -1
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('short54') == True
assert is_acceptable_password('muchlonger') == True
assert is_acceptable_password('ashort') == False
assert is_acceptable_password('muchlonger5') == True
assert is_acceptable_password('sh5') == False
assert is_acceptable_password('1234567') == False
assert is_acceptable_password('12345678910') == True
assert is_acceptable_password('password12345') == False
assert is_acceptable_password('PASSWORD12345') == False
assert is_acceptable_password('pass1234word') == True
print("Coding complete? Click 'Check' to earn cool rewards!")
Sept. 6, 2020
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