Direct solution clear solution for Acceptable Password IV by amandel - python coding challenges

Enable Javascript in your browser and then refresh this page, for a much enhanced experience.

Direct solution solution in Clear category for Acceptable Password IV by amandel

def is_acceptable_password(p):
    return len(p)>5 and (any(x.isdigit() for x in p) and any(not x.isdigit() for x in p)
or len(p)>9)
       
if __name__ == "__main__":
    # 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
    print("Coding complete? Click 'Check' to earn cool rewards!")

July 6, 2022

Comments:

epsilon_pi on June 21, 2023, 12:29 p.m.
Good work!

waffell on Dec. 28, 2023, 9:03 p.m.
You have one mistake, one of the conditions is: "the length should be bigger than 6". If you check a password such as "short5", your function will return True when it should return False. Just replace len( p)>5 with len( p)>6. Apart from that, the solution is good.

amandel on Dec. 28, 2023, 9:08 p.m.
You are right, and that is a silly mistake. I wonder who made the tests - they should have catched this error.

waffell on Dec. 28, 2023, 9:56 p.m.
No one is infallible :-)

Strings Theory

0 %

You ClassRoom Demo is Ready With the new tool CheckiO ClassRoom, you will be able to analyze your students' progress, have your own leaderboard, change missions order on the map, and so on.