Problem Description
Given a valid parentheses string s, consider its primitive decomposition: s = P1 + P2 + ... + Pk, where Pi are primitive valid parentheses strings. Return s after removing the outermost parentheses of every primitive string in the primitive decomposition of s.
Key Insights
- A valid parentheses string is defined by its structure, where each '(' must have a corresponding ')'.
- Primitive strings are those that cannot be split further into valid substrings.
- The outermost parentheses can be identified by counting the balance of parentheses as we traverse the string.
- Removing the outermost parentheses of each primitive can be achieved by keeping track of the depth of nesting.
Space and Time Complexity
Time Complexity: O(n) - we traverse the string once. Space Complexity: O(n) - in the worst case, we store all resulting strings.
Solution
To solve the problem, we can use a counter to track the balance of parentheses. We iterate through the string, and for every character:
- Increment the counter for '(' and decrement for ')'.
- Whenever the counter is at 1 (indicating we're at the outermost level), we skip that '(' and when we reach 0, we skip that ')'. This allows us to build the resulting string without the outermost parentheses.