Two pointers was the right call and I knew it immediately, but I fumbled the duplicate-skipping logic for a bit.
Use a two-pointer technique starting from both ends of the sorted array, moving inward based on the sum compared to the target. To avoid duplicate pairs, skip over duplicate values when moving the pointers after finding a valid pair. This yields O(n) time and O(1) extra space (excluding output).
Pro tip: Explicitly discuss how you handle duplicates to ensure unique pairs, and mention that the output size can be O(n^2) in the worst case, so space complexity should consider the output.
Confirm the problem: sorted array, unique index pairs (i, j) with i < j, sum equals target, no duplicate pairs. Ask if the array can contain duplicates and if the output should be sorted.
Explain that since the array is sorted, two pointers from both ends can efficiently find pairs. Compare the sum of elements at left and right pointers to the target and move pointers accordingly.
When a valid pair is found, add it to the result, then skip all duplicate values on both sides to avoid repeating the same pair. Also, skip duplicates when moving pointers even if no pair is found.
Time complexity: O(n) because each element is visited at most once by each pointer. Space complexity: O(1) extra space, but the output list can take O(n^2) space in the worst case (e.g., many pairs).
Walk through examples: no pairs, all pairs, duplicates, negative numbers. Verify that duplicates are handled correctly and that indices are unique.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.