Enable Javascript in your browser and then refresh this page, for a much enhanced experience.
Four lines solution in Speedy category for Acceptable Password VI by GrigorySol
def is_acceptable_password(password: str) -> bool:
if 'password' in password.lower() or len(set(password)) < 3:
return False
return 0 < sum([i.isdigit() for i in password]) < len(password) > 6 or len(password) > 9
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!")
May 12, 2020
Comments: