I went straight for two pointers since the array is sorted, which is the right call.
Start by clarifying the problem and constraints, then propose the two-pointer technique for the sorted array, explaining its O(n) time and O(1) space complexity. Discuss how to handle follow-ups: for extreme targets, consider integer overflow and edge cases; for unsorted arrays, compare sorting (O(n log n)) vs. hash map (O(n) time, O(n) space) approaches.
Pro tip: Explicitly state that you're tracking the minimum absolute difference and updating the closest pair when a smaller difference is found, and mention that if the target is very large or small, you might need to use a data type with a wider range or handle overflow carefully.
Ask clarifying questions: Is the array sorted? Can it contain duplicates? What should be returned if no pair exists? Are there memory constraints? Confirm the expected output format.
For the sorted array, explain the two-pointer technique: initialize left at 0 and right at n-1, compute sum, update closest pair based on absolute difference, and move pointers inward based on comparison with target.
State time and space complexity: O(n) time, O(1) space. Compare with brute force O(n^2) and binary search O(n log n) approaches, highlighting why two-pointer is optimal for sorted input.
For very large/small targets: discuss integer overflow and potential need for long or arbitrary precision. For unsorted array: compare sorting first (O(n log n) time, O(1) space if in-place) vs. using a hash map (O(n) time, O(n) space).
Walk through a simple example (e.g., array [-1, 2, 3, 4], target=3) to demonstrate correctness, and mention edge cases like empty array, single element, or all elements summing to less than target.
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