Counting while Valid Parentheses

When a pair of valid parentheses is found, just add 2 to count.

However, the counting method is a bit tricky. Show you using animation

Animation

This method adds a counting stack.

input [ ( ( ) ( ) ) ]

count []
stack []
 
       i

No match put ) and ( into stack

input [ ) ( ) ) ]

count [ 0 0 ]
stack [ ( ( ]
 
          i

found matches pop stack and count[1] = count[1] + 2

input [ ( ) ) ]

count [ 0 2 ]
stack [ ( ]
 
        i

just two steps.

found another matches ( and ) pop stack and count[1] = count[1] + 2

input [ ]

count [ 0 4 ]
stack [ ]
 
        i

found matches, but, this time, count[0] = count[0] + 2 is not enough.

It should be count[0] = count[0] + 2 + count[1], because that here 0 matches parentheses indicates that the current parentheses is enclosing the 1..current.

Generally, when meets a match,

count[stack.size()] =  count[stack.size()] + 2 + count[stack.size() + 1]

then reset the count[stack.size() + 1] to 0.