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Still 1 line solution in Clear category for Acceptable Password VI by mAzrunnr
def is_acceptable_password(p: str) -> bool:
return len(p)>6 and len(set(p))>2 and not('password' in p.lower()) and (len(p)>9 or (any(x.isdigit() for x in p) and any(x.isalpha() for x in p)))
June 12, 2021
