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First solution in Clear category for Acceptable Password VI by pakicetus
import re
def is_acceptable_password(password: str) -> bool:
checkOne = not "PASSWORD" in password.upper()
checkTwo = re.search("\d", password) and not password.isnumeric()
checkThree = len(password) > 9
checkFour = len(set(password)) >= 3
checkFive = len(password) > 6
return checkFive and (checkTwo or checkThree) and checkFour and checkOne
July 28, 2020