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once again solution in Clear category for Acceptable Password VI by tom-tom
def is_acceptable_password(pw: str) -> bool:
return ((not all(map(str.isdigit, pw)) and
(len(pw) > 6) and any(map(str.isdigit, pw)) or len(pw) >= 10) and
'password' not in pw.lower() and len(set(pw)) >= 3)
July 13, 2020
