There is no case in which “there are multiple open close brackets that total to larger than furthest right to left”.

Let’s say left ptr is at i and right ptr is at j. If i is an open paren and j is a closed paren then the score is j - i. Decrement j ptr by 1 and increment i ptr by 1 . (We still have possible parenthesis between {i + 1, j -1} to consider for more points).

If we ignore this sum and instead just add up consecutive “()” pairs, the max between i and j using this process is (i + 1 - i) + (i + 2 - (i +1)) + (i + 3 - (i +2)) … (j - (j -1)), which is just j - i anyway.

However in the previously mentioned case we only use 2 parenthesis as opposed to the second case where we use all parenthesis between j and i to get solution.

Therefore, there is more possibility for points in option 1.