Part A felt easy and I went straight to a counter, just incrementing on '(' and decrementing on ')'.
Start by clarifying the problem and edge cases, then propose a stack-based solution for parentheses. Walk through the algorithm step-by-step, and extend it to handle all three bracket types by using a mapping of closing to opening brackets. Analyze time and space complexity, and discuss potential optimizations or alternative approaches.
Pro tip: Mention that you would use a stack and a hash map for matching brackets, and emphasize that this approach naturally handles nested and interleaved brackets. Also, proactively discuss edge cases like empty strings and strings with only closing brackets.
Ask if the string can be empty, if it contains only brackets, and if we need to handle all three types. Confirm that validity means every opening bracket is closed by the same type in the correct order.
Explain that a stack is ideal because brackets must be closed in LIFO order. For parentheses only, push opening brackets and pop when a closing bracket is encountered, checking for emptiness.
Introduce a mapping (e.g., hash map) from closing to opening brackets. When a closing bracket is seen, check if the stack is non-empty and the top matches the expected opening bracket.
Trace the algorithm on a sample string like '{[()]}' to demonstrate correctness, and also on an invalid string like '([)]' to show how mismatches are caught.
State that time complexity is O(n) and space complexity is O(n) in the worst case. Mention that this is optimal for a single-pass solution, and briefly discuss alternatives like using a counter for parentheses only (but not for multiple types).
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