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Acceptable Lambda solution in Clear category for Acceptable Password VI by cterrazas2
short_or_long = lambda password: (len(password) > 6 and any(x.isdigit() for x in password) and any(x.isalpha() for x in password)) if len(password) < 10 else True
no_password = lambda password: 'password' not in password.lower()
three_diff = lambda password: len(set(password)) >= 3
def is_acceptable_password(password: str) -> bool:
return short_or_long(password) and no_password(password) and three_diff(password)
if __name__ == '__main__':
print("Example:")
print(is_acceptable_password('short'))
# These "asserts" are used for self-checking and not for an auto-testing
assert is_acceptable_password('short') == False
assert is_acceptable_password('short54') == True
assert is_acceptable_password('muchlonger') == True
assert is_acceptable_password('ashort') == False
assert is_acceptable_password('muchlonger5') == True
assert is_acceptable_password('sh5') == False
assert is_acceptable_password('1234567') == False
assert is_acceptable_password('12345678910') == True
assert is_acceptable_password('password12345') == False
assert is_acceptable_password('PASSWORD12345') == False
assert is_acceptable_password('pass1234word') == True
assert is_acceptable_password('aaaaaa1') == False
assert is_acceptable_password('aaaaaabbbbb') == False
assert is_acceptable_password('aaaaaabb1') == True
assert is_acceptable_password('abc1') == False
assert is_acceptable_password('abbcc12') == True
print("Coding complete? Click 'Check' to earn cool rewards!")
July 4, 2020
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