First clear solution for Acceptable Password IV by StanislauL - python coding challenges

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First solution in Clear category for Acceptable Password IV by StanislauL

def is_acceptable_password(password: str) -> bool:
    return (len(password)>6 and not password.isdigit() and not password.isalpha()) or len(password)>9

Aug. 24, 2020

Comments:

Mikhail_Podgorny on Jan. 18, 2021, 4 p.m.
Solution doesn't work on this test string 'this 1t' I think better change "not password.isalpha()" on "any(sym.isdigit() for sym in password)".

jakubbortlik on Nov. 30, 2022, 2:53 p.m.
I don't find this solution very "clear", after all the conditions have been combined to one line.

raj0vaibhav on Dec. 31, 2022, 6:23 a.m.
<h1>This is failing</h1> <h1>Problem statement is only about Alphanumeric passwords</h1> assert isacceptablepassword('_9') == False

Germ on March 10, 2023, 3:10 p.m.
This doesn't work. I tried using the password "_" and it took it as valid. In general, this has a chance to fail whenever the password has something that is not a number or a letter: underscores, spaces, anything else. Because it doesn't check if there is at least a number, it checks if there is something that is not a letter.

Strings Theory

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