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Find 'password' solution in Clear category for Acceptable Password V by Striga
def is_acceptable_password(password: str) -> bool:
if password.lower().find('password') != -1: return False
if len(password) > 9: return True
return 6 < len(password) != len([x for x in password if x.isdigit()]) > 0
if __name__ == '__main__':
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
assert is_acceptable_password("this is password") == False
Oct. 15, 2020
