Solution
Approach 1: Divide and Conquer
Intuition
Split the string into S = A + B
where A
and B
are balanced parentheses strings, and A
is the smallest possible nonempty prefix of S
.
Algorithm
Call a balanced string primitive if it cannot be partitioned into two nonempty balanced strings.
By keeping track of balance
(the number of (
parentheses minus the number of )
parentheses), we can partition S
into primitive substrings S = P_1 + P_2 + ... + P_n
. Then, score(S) = score(P_1) + score(P_2) + ... + score(P_n)
, by definition.
For each primitive substring (S[i], S[i+1], ..., S[k])
, if the string is length 2, then the score of this string is 1. Otherwise, it's twice the score of the substring (S[i+1], S[i+2], ..., S[k1])
.
Complexity Analysis

Time Complexity: , where is the length of
S
. An example worst case is(((((((....)))))))
. 
Space Complexity: , the size of the implied call stack.
Approach 2: Stack
Intuition and Algorithm
Every position in the string has a depth  some number of matching parentheses surrounding it. For example, the dot in (()(.()))
has depth 2, because of these parentheses: (__(.__))
Our goal is to maintain the score at the current depth we are on. When we see an opening bracket, we increase our depth, and our score at the new depth is 0. When we see a closing bracket, we add twice the score of the previous deeper part  except when counting ()
, which has a score of 1.
For example, when counting (()(()))
, our stack will look like this:
[0, 0]
after parsing(
[0, 0, 0]
after(
[0, 1]
after)
[0, 1, 0]
after(
[0, 1, 0, 0]
after(
[0, 1, 1]
after)
[0, 3]
after)
[6]
after)
Complexity Analysis

Time Complexity: , where is the length of
S
. 
Space Complexity: , the size of the stack.
Approach 3: Count Cores
Intuition
The final sum will be a sum of powers of 2, as every core (a substring ()
, with score 1) will have it's score multiplied by 2 for each exterior set of parentheses that contains that core.
Algorithm
Keep track of the balance
of the string, as defined in Approach #1. For every )
that immediately follows a (
, the answer is 1 << balance
, as balance
is the number of exterior set of parentheses that contains this core.
Complexity Analysis

Time Complexity: , where is the length of
S
. 
Space Complexity: .