Plesae tell what is the approach to solve Brackets all over problem

#1

Not able to build logic.

#3

can you please explain this:
or add main string if ( open >= pre and taken == false ) then make open = open — pre + bef

#4

pre is the minimum no of open brackets that is needed before main string to make it valid.

so

 if ( open >= pre and taken == false )

this if is checking whether we can place main string or not ie do we have sufficient number of open brackets (open >= pre) and is string availabel (taken==false).
if both condition is valid then we can place our main string

now coming to
open=open-pre+bef

pre is the open brackets required before main string. so when we add main string these open brackets will get paired with closed bracket present in main string hence we need to remove the paired brackets(pre) from the count so latest open bracket count =open-pre .
and now becuase we are required to put bef number of closed brackets after main string that means this string has bef number of extra open brackets that why we need add this extra open brackets in our open count
which is why the final open bracket counts=
open=open -pre + bef

#5

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