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Simple solution in Clear category for Acceptable Password VI by thealfest1
def is_acceptable_password(password: str) -> bool:
return len(set(password)) > 2 and (len(password) > 9
or (0 < sum(char.isdigit() for char in password) < len(password) > 6)
) and not 'password' in password.casefold()
March 19, 2020
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