Instead of replacing the brackets, you could do just one loop, and keep a stack with the opening brackets. Every time you find a closing bracket, check if it corresponds to the stack top: if it's not, the string is invalid, else, pop the opening bracket and proceed to the next character.

Do it until the end of the string, and the stack will be empty if it's a valid one:

def is_balanced(s, brackets={ '(': ')', '[': ']', '{': '}' }):
    if len(s) % 2 != 0:
        return False

    open_brackets = set(brackets.keys())
    close_brackets = set(brackets.values())
    stack = []
    for c in s:
        if c in open_brackets:
            stack.append(c)
        elif c in close_brackets:
            # close bracket without respective opening, string is invalid
            if len(stack) == 0:
                return False
            # close bracket doesn't correspond to last open bracket, string is invalid
            if c != brackets[stack[-1]]:
                return False
            stack.pop()
        else: # invalid char
            return False

    return len(stack) == 0

I put the keys and values inside set's (because searching with the in operator is O(1)), but it didn't make much difference. You can remove it if you want.

I admit I initially thought that calling replace 3 times per iteration, and always do it N / 2 times, would be slower, but to my surprise I found out that it depends on the string being tested. A quick test with timeit module (using your isValid function and is_balanced as defined above):

from timeit import timeit

# run 10 times each test
params = { 'number' : 10, 'globals': globals() }

s = '{[(())]{()[()]}}' * 1000
s += '))'
s += '{[(())]{()[()]}}' * 1000
print('--------------\nUnbalanced in the middle')
print(timeit('isValid(s)', **params))
print(timeit('is_balanced(s)', **params))

s = '))'
s += '{[(())]{()[()]}}' * 1000
s += '{[(())]{()[()]}}' * 1000
print('--------------\nUnbalanced in the beginning')
print(timeit('isValid(s)', **params))
print(timeit('is_balanced(s)', **params))

s = '{[(())]{()[()]}}' * 2000
s += '))'
print('--------------\nUnbalanced in the end')
print(timeit('isValid(s)', **params))
print(timeit('is_balanced(s)', **params))

s = '{[(())]{()[()]}}' * 2000
print('--------------\nBalanced')
print(timeit('isValid(s)', **params))
print(timeit('is_balanced(s)', **params))

In my machine, I've got these results:

--------------
Unbalanced in the middle
0.03587988700019196
0.02324946300359443
--------------
Unbalanced in the beginning
0.026359478004451375
1.2465003237593919e-05
--------------
Unbalanced in the end
0.025080425999476574
0.041097565001109615
--------------
Balanced
0.022329219995299354
0.04195229199831374

For a balanced string, surprisingly - at least to me - the replace approach is faster.

For unbalanced strings, it depends on where the unbalanced brackets are. If they are in the beginning or in the middle, the stack approach wins (with a huge margin if the problem is in the beginning, due to the else clause that checks an invalid char as soon as it finds one). But if they are in the end, the replace approach wins.

I really thought that using replace would always be slower, because each call to replace needs to scan the whole string, searching for the substring to be replaced (and return a new string). Even if you're looping N / 2 times, you're calling replace 3 times per iteration, which I really thought it'd have a huge cost. But that wasn't the case, and this really surprised me.

The only case where the difference is really huge is the aforementioned case, when the unbalanced bracket is in the beginning. One particular case is curious:

s = '{[((ab))]{()[()]}}' * 2000

With this, the replace approach took 4.5 seconds, while the stack approach took around 0.00001 seconds. Not sure if there's an implementation detail for the replace method, that makes this particular case very slow (tested on Python 3.8.5, Ubuntu 20.04).

Another case where the stack approach wins by one order of magnitude is:

# lots of opening brackets, followed by lots of closing brackets
s = ('(' * 2000) + (')' * 2000)

But as I said, I was expecting this difference in all cases. Anyway, that's why we should always test everything, instead of trusting on what we believe to be "obvious".

BTW, testing in another environments, the results vary. In IdeOne.com, the stack approach is always faster for unbalanced strings, and almost the same for balanced ones. And in Repl.it, the results were similar to my machine's.

posted over 3 years ago

CC BY-NC-SA 4.0

3y ago

hkotsubo‭

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