Classic backtracking problem and I knew it, which helped.
Use a backtracking algorithm that builds the string incrementally, adding an opening parenthesis if we haven't used all n, and a closing parenthesis if the number of closing parentheses is less than opening. This ensures all generated strings are valid and avoids unnecessary recursion.
Pro tip: Mention that the number of valid combinations is the nth Catalan number, and that the time complexity is O(4^n / sqrt(n)) or simply O(C_n * n) where C_n is the Catalan number. This shows deeper understanding and awareness of combinatorial limits.
Confirm that n is a non-negative integer and that the output should be a list of strings. Discuss edge cases like n=0 (should return [''] or []) and n=1 (should return ['()']).
Explain that backtracking is ideal because we can prune invalid paths early. Maintain counts of open and close parentheses used so far.
At each step, add '(' if open < n, and add ')' if close < open. Recurse with updated counts and current string. When the string length reaches 2n, add it to the result.
State that the number of valid combinations is the nth Catalan number, and the time complexity is O(4^n / sqrt(n)) or O(C_n * n). Space complexity is O(n) for recursion depth plus output storage.
Walk through a small example like n=2 to verify correctness. Mention that using a StringBuilder or list of characters can improve performance over string concatenation.
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