Score of Balanced Parentheses

Given a string s, consisting of pairs of balanced parentheses, find the score of the given string based on the given rules:

• “()” has a score of 1.
• “A B” has a score of A + B, where A and B are individual pairs of balanced parentheses.
• “(A)” has a score twice of A i.e., the score is 2 * score of A.

Examples:

Input: s = “()()” 
Output: 2 
Explanation: The string s is of the form "AB", that makes the total score = (score of A) + (score of B) = 1 + 1 = 2.

Input: s = "(()(()))"
Output: 6
Explanation: The string s is of the form "(A(B))" which makes the total score = 2 × ((score of A) + 2 × (score of B)) = 2 × (1 + 2 × (1)) = 6.

[Approach] Using Stack - O(n) Time and O(n) Space

Think of the string as a tree-like structure:

• Every '(' means we are going down to a child level.
• Every ')' means we are returning back to the parent level.

To keep track of scores at each level, we use a stack.

Now, how does the stack help?

The stack keeps track of the score at each level. When we go down '(', we push 0 to start for that level. When we come back through ')', we pop the score of that level:

• If it was empty "()", score = 1.
• If it had children "(A)", score = 2 × innerScore.

Then we add this score to the parent’s score (the new top of the stack). At the end, only the root score is left in the stack and that’s our total score.

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