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acceptable_password_v solution in Clear category for Acceptable Password V by dannedved
def is_acceptable_password(password: str) -> bool:
digit_count = sum([char.isdigit() for char in password])
find_pass = password.lower().find("password")
return find_pass == -1 and (len(password) > 9 or len(password) > 6 and digit_count in range(1, len(password)))
April 27, 2020
