Knew it was backtracking pretty fast but fumbled the base case the first time around.
Use a backtracking algorithm that builds the string incrementally, ensuring at each step the number of closing parentheses does not exceed opening ones. Recursively explore adding an opening parenthesis if available, and a closing parenthesis if it won't invalidate the sequence. Collect results when the string length reaches 2n.
Pro tip: Mention that this is a classic example of backtracking with pruning, and that the number of valid combinations is the nth Catalan number, which shows you understand the combinatorial complexity. Also, clarify that the solution generates only valid sequences, avoiding the need to filter invalid ones.
Confirm that n is a positive integer and that the output should be a list of strings representing all valid combinations. Discuss edge cases like n=0 (should return an empty list or a list with an empty string, depending on definition).
Decide on parameters: current string, number of open parentheses used, and number of close parentheses used. The goal is to reach open == close == n.
When the current string length equals 2n, add it to the result. Otherwise, try adding '(' if open < n, and try adding ')' if close < open.
Write the recursive function, ensuring to backtrack by removing the last character after each recursive call. Analyze time complexity as O(4^n / sqrt(n)) or O(C_n) where C_n is the nth Catalan number, and space complexity as O(n) for recursion depth plus output storage.
Walk through n=3 to show the generation process. Mention that the algorithm is optimal as it only generates valid sequences, and discuss potential iterative solutions or dynamic programming if asked.
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