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Concise algorithm solution in Speedy category for Acceptable Password VI by Igor_Sekretarev
def is_acceptable_password(password: str) -> bool:
return ( 'password' not in password.casefold()
and len(set(password)) > 2
and (len(password) > 9 or ( len(password) > 6
and any(ch.isdigit() for ch in password)
and not password.isdigit())))
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!")
April 25, 2021
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