The recursive backtracking approach is what they're looking for: track how many open and close parens you've placed, and only add a closing paren when the close count is less than the open count.
Use backtracking to build valid parentheses strings incrementally, ensuring at each step the number of closing parentheses never exceeds opening ones. Track the counts of open and close parentheses used, and only add '(' if open < n, and ')' if close < open. When the string length reaches 2n, add it to the result.
Pro tip: Discuss the time complexity as O(4^n / sqrt(n)) (the nth Catalan number) and mention that this is optimal since you must generate each valid string. Also, briefly note how this combinatorial generation could be relevant to ML tasks like generating structured outputs or parsing expressions.
Confirm that n is a non-negative integer and that the output should be a list of strings. Define what constitutes a well-formed parentheses string: equal number of '(' and ')', and at any prefix, the number of ')' does not exceed '('.
Explain that backtracking is ideal because it explores all valid combinations by making choices and undoing them. It prunes invalid paths early, avoiding unnecessary work.
Describe the recursive function: parameters are the current string, count of open parentheses, and count of close parentheses. At each call, if open < n, add '(' and recurse; if close < open, add ')' and recurse. Base case: when string length == 2n, add to result.
State that the number of valid combinations is the nth Catalan number, so time complexity is O(4^n / sqrt(n)) and space complexity is O(n) for recursion depth (excluding output storage).
Walk through n=3 to show how the algorithm generates all 5 valid strings, ensuring no duplicates and all are well-formed.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.