14 ms·
Yeah, that's the first solution that popped in my mind too. I think it needs just a special case at the beginning to check if the first paren is not a closing o
by Epholys 8y ago
Yeah, that's the first solution that popped in my mind too. I think it needs just a special case at the beginning to check if the first paren is not a closing one.
EDIT: yep, after reading the comments below I see that I really should have put just a little more thought before dismissing this as too trivial to think about.
- skybrian 8y agoA test case: ())(
- FreeFull 8y agoWhat you actually need is to check if the count ever goes negative.
- braythwayt 8y agoThe stack solution checks that the stack is non-empty before popping. The counter solution simply needs to check that the counter is non-zero before decrementing. All cases then work perfectly.