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First solution in Clear category for Acceptable Password V by makoto_yamagata
def is_acceptable_password(password: str) -> bool:
any_ = [
lambda a: len(a) > 9,
lambda a: not a.isdigit() and len(a) > 6 and a[-1].isdigit()
]
all_ = [
lambda a: "password" not in a.lower(),
lambda a: any(map(lambda f: f(a), any_)),
]
return all(map(lambda f: f(password), all_))
Dec. 26, 2020
