[Epholys]

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.

[braythwayt]

The 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.

[FreeFull]

What you actually need is to check if the count ever goes negative.

[skybrian]

A test case: ())(