Score of Parentheses - Solution & Explanation

This approach leverages a stack to manage the scores of the balanced parentheses string. As we iterate through the string:

• Push 0 onto the stack for every '('.
• For every ')', pop the top element (score) from the stack and update it. If the popped score is zero, it means it was a simple pair '()', so the score is at least 1. Then, double this score if necessary and add it to the current top of the stack.

At the end of the iteration, the top of the stack will contain the total score.

The C solution utilizes a simple stack array to keep track of scores at each level. As we parse the string, we push zero onto the stack for each '('. When encountering a ')', the score is calculated and added to the top of the stack. This method ensures that nested structures are calculated accurately by collapsing scores upwards through multiplication and addition.

This approach iteratively computes the score by tracking the depth of parentheses:

• Keep an array or list where the index indicates depth and the value holds the score.
• Increment the depth for '(', decrease it for ')'.
• On encountering ')', when at zero depth, update the previous depth's score. If it completes a '()', increment the current score; if it ends a nested structure, double the score.
• The total score is summed by the end.

In C, this iterative approach uses an array to track scores at different depths. It handles the transition from '(' to ')' by either recognizing single pairs or multiplying for nested structures. This approach exploits direct index manipulation for score calculations.