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First solution in Speedy category for Acceptable Password VI by _Chico_
import re
def is_acceptable_password(password: str) -> bool:
return bool(re.match(r"""
((?=.*\d)(?=.*\D)(?=.{7,})|(?=.{10,}))
(?!(?i:.*password))
(?P.).*?(?!(?P=x))(?P.).*?(?!(?P=x))(?!(?P=y))(.)""",
password, re.VERBOSE))
May 23, 2021
