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First solution in Speedy category for Acceptable Password VI by Matt_Leroy
def is_acceptable_password(x):
return (
(len(x) > 9 or len(x) > 6 and not x.isdigit() and not x.isalpha())
and "password" not in x.lower()
and len(set(x)) >= 3
)
May 22, 2023
