def scoreOfParentheses(s):
    """
    :type s: str
    :rtype: int
    """
    # Attempted Solution
    # score = left = right = 0

    # for x in s:
    #     if x == "(":
    #         left += 1
    #         score += 1
    #     elif x == ")":
    #         right += 1
    #         score += 1
    # if (left + right) % 2 == 0:
    #     score /= 2
    # return score

    # 96% faster solution
    # Initialize an empty list 'a', the length of the input string 's' as 'n'
    a, n = [], len(s)
    
    # Initialize 'c' and 't' to 0.
    c = t = 0
    
    # Loop through the characters of the input string, starting from the second character (index 1).
    for i in range(1, n):
        print(f'c is {c}')
        print(t)
        # If the current character is an opening parenthesis '('
        if s[i] == '(':
            # Increment the temporary variable 't' to keep track of the depth of nesting.
            t += 1
        # If the current character is a closing parenthesis ')', and the previous character was an opening parenthesis '('
        elif s[i - 1] == '(':
            # Calculate the score for the current balanced pair of parentheses and add it to 'c'.
            # The score is 2 raised to the power of 't' (2^t).
            c += 1 << t
            
            # Decrease the temporary variable 't' as the current nested pair is closed.
            t -= 1
        else:
            # If the current character is a closing parenthesis ')' and the previous character was also a closing parenthesis ')',
            # simply decrease the temporary variable 't' to indicate the decrease in the depth of nesting.
            t -= 1
    
    # Return the total score of the balanced parentheses expression, stored in 'c'.
    return c

# print(scoreOfParentheses("()")) # 1
print(scoreOfParentheses("(()(()))")) # 6
# print(scoreOfParentheses("()()()")) # 3