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Second solution in Clear category for Acceptable Password VI by tigelane
def is_acceptable_password(password: str) -> bool:
if "password" in password.lower():
return False
if len(set(password)) < 3:
return False
if len(password) >= 9:
return True
if len(password) <= 6:
return False
if has_numbers(password) == False:
return False
if password.isdigit():
return False
return True
def has_numbers(password):
# Has at least 1 number
return any(char.isdigit() for char in password)
Sept. 28, 2020
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