The other answers have already covered what can be improved in your script so I will not repeat any of that. Just wanted to add an alternative I found interesting.
An elimination-based approach:
def f(my_string):
brackets = ['()', '{}', '[]']
while any(x in my_string for x in brackets):
for br in brackets:
my_string = my_string.replace(br, '')
return len(my_string) > 0
In every iteration the innermost brackets get eliminated (replaced with empty string). If we end up with an empty string, our initial one was balanced; otherwise, not.
Examples and notes:
- Best case:
s = '[](){}{}()[]'-> reduces to nothing in onewhileiteration. - Worst case:
s = '({[[{()}]]})'-> same length string requires 6 iterations (destroyed inside out)
You could add some short-circuiting for quickly catching cases of lengthy strings that have an uneven number of opening and closing brackets so that you don't waste your time eliminating...
