Generate Parentheses
Given $n$ pairs of parentheses, write a function to generate all combinations of well-formed parentheses.
Why Interviewers Ask This
Google asks this question to evaluate a candidate's mastery of recursion, backtracking, and state management. It tests the ability to prune invalid search spaces efficiently rather than generating all permutations blindly. Interviewers specifically look for logical clarity in maintaining balance constraints while constructing valid sequences dynamically.
How to Answer This Question
1. Clarify Requirements: Confirm input n is positive and define 'well-formed' as every opening bracket having a matching closing bracket in correct order. 2. Define State Variables: Identify that you need to track the count of open brackets used and closed brackets placed so far. 3. Establish Base Case: Determine when to stop recursion, which occurs when the current string length equals 2*n. 4. Formulate Recursive Logic: Explain that you can add an open parenthesis if open_count < n, and a closing parenthesis only if closed_count < open_count. 5. Pruning Strategy: Emphasize how these conditions naturally prevent invalid combinations without needing post-generation validation. 6. Complexity Analysis: Discuss time complexity O(4^n / sqrt(n)) based on Catalan numbers and space complexity O(n) for the recursion stack depth.
Key Points to Cover
- Explicitly defining the pruning condition (close_count < open_count) to avoid invalid states
- Demonstrating understanding of Catalan numbers regarding the output size
- Clearly articulating the base case where string length equals 2*n
- Explaining why brute force generation followed by filtering is inefficient
- Maintaining clean variable naming and logical flow in the explanation
Sample Answer
To solve the Generate Parentheses problem efficiently, I would use a backtracking approach with pruning. First, I'd clarify that we need exactly n pairs where every opening bracket has a corresponding closing one. My sol…
Common Mistakes to Avoid
- Attempting to generate all 2^(2n) permutations first and then filtering them out, which is computationally wasteful
- Failing to pass the current counts or string by reference/value correctly, leading to shared state bugs
- Not handling edge cases like n=0 or negative inputs before starting the recursion
- Confusing the condition for adding a closing bracket, such as checking against n instead of open_count
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