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First: set solution in Clear category for Acceptable Password VI by leggewie
# the logic is getting a bit more complicated now
# test for "password" and minimum amount of different characters first for fast failure
def is_acceptable_password(password: str) -> bool:
return not "password" in password.lower() and \
len(set(password)) > 2 and \
(len(password) > 9 or \
(len(password) > 6 and \
any(char.isdigit() for char in password) \
and not password.isdigit()))
June 9, 2021
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