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First solution in Clear category for Acceptable Password VI by Oleg_Domokeev
def is_acceptable_password(password: str) -> bool:
return (len(password) > 9 or (len(password) > 6 and any(sym.isdigit() for sym in password) and not password.isdigit())) and password.lower().find('password') == -1 and len(set(password)) > 2
July 13, 2020
