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set(iterable) solution in Clear category for Acceptable Password VI by David_Jones
def is_acceptable_password(password):
return ( 'password' not in password.lower()
and len(set(password)) > 2
and (len(password) > 9 or len(password) > 6
and any(ch.isdigit() for ch in password)
and not password.isdigit()))
March 19, 2020
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