First uncategorized solution for House Password by Or123 - python coding challenges

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First solution in Uncategorized category for House Password by Or123

def checkio(data):
    'Return True if password strong and False if not'
    num = False
    upper = False
    lower = False
    for i in range(len(data)):
        num = num or data[i].isnumeric()
        upper = upper or data[i].isupper()
        lower = lower or data[i].islower()
        
    return num and upper and lower and len(data)>=10

if __name__ == '__main__':
    assert checkio('A1213pokl')==False, 'First'
    assert checkio('bAse730onE4')==True, 'Second'
    print ("All ok")

May 28, 2012

Comments:

Tommey on Nov. 11, 2012, 12:49 p.m.
You should check for the length minimum requirement in the beginning.

Or123 on Nov. 16, 2012, 4:22 p.m.
I'll leave it that way here, but when I implement it anywhere, I'll be sure to do that, will definitely save calculation time. Thanks! :)

Finaldes on Nov. 19, 2012, 6:08 a.m.
easy

r_geoghegan on Dec. 8, 2012, 10:07 p.m.
On one hand, it is inefficient because you have to iterate through every character to check the password, even if the first three fulfill the conditions. However, it is a much more secure way to do it because you avoid timing attacks! <a href="http://en.wikipedia.org/wiki/Timing_attack">http://en.wikipedia.org/wiki/Timing_attack</a>

justatest90 on Dec. 12, 2012, 11:01 p.m.
How is this the #1 solution?

bryukh on April 17, 2013, 8:51 p.m.
People voted up.

guido on April 16, 2013, 12:43 a.m.
<pre class='brush: python'> 'Return True if password strong and False if not' </pre> Most programming style guides use "triple double" quotes for docstrings, e.g. <pre class='brush: python'>"""Return True if the password is strong and False if not.""" </pre> I also recommend using full grammatical sentences and ending with a period.

guido on April 16, 2013, 12:44 a.m.
This is a perfectly acceptable solution for passwords, which are typically not very long. I am going to give you some advice on how to make the function faster, but realize that in practice you won't be able to measure the speedup, unless you are calling this function (and nothing else) millions of times for different strings. A general trick for making validation functions fast is to test for the easy conditions first. In this case, if the data is too short, you'd be wasting time in the for-loop. In Python you are allowed to have multiple return statements per function, so you could add this to the top of your function: <pre class='brush: python'>if len(data) < 10: return False </pre> Then you can replace the final return with <pre class='brush: python'>return num and upper and lower </pre> Next, you are not using all the conveniences of the for-loop! Instead of writing <pre class='brush: python'>for i in range(len(data)): </pre> and then using data[i] in the body of the loop three times, you can iterate over the characters of the input string directoy: <pre class='brush: python'>for char in data: </pre> and then in the body, use char instead of data[i]. Finally, but this is debatable, once you have found one character in each category, you can stop looking for further characters in that category. You are sort of doing this already by using the shortcut 'or' operator. However, even if the first character is a number, the second is uppercase, and the third character is lowercase, you would still be going through the loop until the end of the string, even though this doesn't add any more information. Consider adding this to the end of the loop: <pre class='brush: python'> if num and upper and lower: return True </pre> Now you can replace the final return statement with simply <pre class='brush: python'>return False </pre>

skeet70 on April 17, 2013, 8:08 p.m.
Wow. Tips from Guido, score Or123!

guido on April 16, 2013, 12:45 a.m.
<pre class='brush: python'>if __name__ == '__main__': </pre> It's hardly necessary to add the 'First' or 'Second' messages to the assert statements, as it prints the failed line in the traceback, so you can see exactly which line went wrong already. Adding a message to assert is useful (a) when there is a framework that suppresses the traceback, or (b) when the programmer would need more information, e.g. <pre class='brush: python'>assert actual == expected, (actual, expected) </pre> Also note that the message doesn't have to be a string -- anything else will be stringified by calling str() on it.

Or123 on June 5, 2013, 11:29 a.m.
<pre class='brush: python'>if __name__ == '__main__': </pre> Thank you very much for all the useful tips, I will try to use them in the next code implementations.

xclam on April 16, 2013, 1:05 p.m.
<pre class='brush: python'> lower = False </pre> why don't use a regular expression?

Ganet on April 20, 2013, 5:43 a.m.
I agree with guido's comment about the loop iteration and testing the length of the string first of all. On the other hand the required password length is so small that testing the length before won't improve the performance reasonably.

zn01wr on April 26, 2013, 4:50 p.m.
<pre class='brush: python'> for i in range(len(data)): </pre> You should use "for c in data"

MorphiusFaydal on May 6, 2013, 7:59 a.m.
Looks good. Short, simple and to the point.

oduvan on June 5, 2013, 9:34 a.m.
ok

TarasBasiuk on Oct. 31, 2013, 1:09 a.m.
<pre class='brush: python'> num = num or data[i].isnumeric() </pre> That's clever ;)

TarasBasiuk on Oct. 31, 2013, 1:11 a.m.
<pre class='brush: python'> return num and upper and lower and len(data)>=10 </pre> Check for length from the start, no need to check anything else if it fails.

Reloader on Nov. 13, 2013, 2:39 a.m.
Marvelous!

forrest1988 on Dec. 10, 2013, 1:34 p.m.
Nice solution

VolodymyrSydoruk on Dec. 16, 2013, 5:19 p.m.
Good and clean

qzshadow on Feb. 20, 2015, 7:50 a.m.
a better way to iterate 'data', I think, you could use for c in data: num = num or c.isdigit() ... directly rather than using range and index[].

Strings Theory

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