I knew this was a backtracking problem pretty quickly, the trick is tracking how many open and close brackets you've placed so far and only adding a close bracket when it won't break validity.
Use a backtracking algorithm that builds the parentheses string incrementally, ensuring at each step that the number of closing parentheses never exceeds the number of opening ones. Recursively explore adding an opening parenthesis if we haven't used all n, and a closing parenthesis if it's valid, until the string length reaches 2n.
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 exponential, but the backtracking prunes invalid branches early. Also, clarify that the output size itself is Catalan(n), so any algorithm must take at least that much time.
Confirm that n is a non-negative integer and that the output should be a list of all valid strings. Ask if n can be 0 (should return an empty list or a list with an empty string? Typically, for n=0, return an empty list).
Decide to use recursion with backtracking, maintaining counts of open and close parentheses used so far. Alternatively, you could use a BFS/DFS with a queue, but backtracking is more space-efficient.
Write a helper function that takes the current string, number of open parentheses, and number of close parentheses. The base case is when the string length equals 2*n; add it to the result.
If open < n, add an opening parenthesis and recurse. If close < open, add a closing parenthesis and recurse. This ensures only valid combinations are generated.
Explain that the time complexity is O(4^n / sqrt(n)) due to the Catalan number of outputs, and space complexity is O(n) for recursion depth. Test with n=0, n=1, n=2, and n=3.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.