First uncategorized solution for Brackets by Avaris - python coding challenges

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First solution in Uncategorized category for Brackets by Avaris

# migrated from python 2.7
def checkio(Expression):
    'Check expression for correct brackets order'
    x = "".join(a for a in Expression if a in "{}()[]")
    while ("()" in x) or ("[]" in x) or ("{}" in x):
        x=x.replace("()","")
        x=x.replace("{}","")
        x=x.replace("[]","")
    return len(x)==0

if __name__ == '__main__':
    assert checkio('({(asda)sd[s]d})') == True, 'First'
    assert checkio('({[a]((s)})') == False, 'Second'
    print('All ok')

May 1, 2011

Comments:

old_user on May 17, 2011, 7:48 a.m.
Intresting solution.

limon on May 17, 2011, 1:26 p.m.
so simple! thanks!

Lom on Oct. 24, 2011, 9:46 a.m.
This solution, for example, will not give error position if needed, so I dislike it.

forlogin on March 6, 2012, 7:19 a.m.
A new Solution! wonderful!

Avaris on March 9, 2012, 2:17 a.m.
Thanks all :). BTW, I know all the problems with this code. In addition to point Lom raised, it is much slower than the "usual" way. But I couldn't resist the simplicity of it :).

mayao11 on April 8, 2012, 1:34 a.m.
Slow but very interesting!

veky on May 29, 2013, 7:40 p.m.
Really nice solution.

LinqiXin on June 10, 2013, 4:27 a.m.
Simple Solution!

bylijinnan on June 29, 2013, 2:39 a.m.
Nice although a little slow

miklasiu on July 3, 2013, 12:27 p.m.
Better than mine ;)

zoharek on July 9, 2013, 9:05 p.m.
Regardless of performance, looks interesting, and is very readable.

jkereako on July 12, 2013, 10:23 p.m.
This is way better than mine. Nicely done.

mr147258369 on Sept. 2, 2013, 3:17 a.m.
Very nice solution!

sblight13 on Sept. 9, 2013, 5:14 p.m.
<pre class='brush: python'> return len(x)==0 </pre> elegant!

nickie on Nov. 17, 2013, 2:56 p.m.
This requires O(N^2) time and has to go all the way. Sorry, but I don't think it should be the top voted solution.

Candy_GHL on Nov. 25, 2013, 7:15 a.m.
Amazing!!

ysenko on Dec. 4, 2013, 10:54 a.m.
Complexity is O(N^2). You can write less elegant code but with complexity O(N). Sorry I cannot say that it's the best solution.

wccrawford on Dec. 11, 2013, 6:47 p.m.
Nicely done.

ultras on Dec. 17, 2013, 12:48 p.m.
The in operators can be easily omitted. The loop can be stopped when the len or the value of x is not changed anymore after executing replace.

shreekar on Dec. 26, 2013, 6:30 a.m.
neat, but I think it is not very efficient for very long expressions with lots of brackets.

ignalion on Jan. 18, 2014, 1:08 p.m.
<ol> <li>More than one pass through the expr</li> <li>Hardcoded brackets</li> <li>Replace method on a string is very inefficient here.</li> </ol> Thus, I don't understand the votes.

robertabegg on Sept. 27, 2021, 7:47 p.m.
I have to agree that the performance concerns far outweigh the "interesting" factor here. Ends up creating a lot of string objects also.

hbczmxy on Oct. 13, 2021, 11:58 a.m.
<pre class='brush: python'>while conditon: while-block </pre> if condition is True, it execute while-block if condition is False, skip the while loop and execute the statement behind the loop

lechrudik on May 14, 2022, 6:55 a.m.
elegant and very clear. Even for a beginner. Good job

Strings Theory

0 %

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