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First solution in Clear category for Acceptable Password VI by logicalladybuglifestyle
def is_acceptable_password(password: str) -> bool:
if 'password' in password.lower() or len(set(password)) <= 2:
return False
check = sum([i.isnumeric()for i in password])
l = len(password)
return (l > 9) or (l > 6 and check and check != l)
April 4, 2020
