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First solution in Clear category for Acceptable Password VI by martin_b
def is_acceptable_password(p: str) -> bool:
return len(set(p))>2 and not 'password' in p.lower() and (len(p) > 9 or len(p) > 6 and any(i.isdigit() for i in p) and not all(i.isdigit() for i in p))
Dec. 27, 2020
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