My first instinct was brute force all pairs and sort them, which works but blows up on large inputs.
Clarify the problem constraints (e.g., whether arrays can contain duplicates, if k can exceed total pairs, and if the output should be sorted). Then propose a min-heap approach that starts with the smallest sum (0,0) and expands to neighbors (i+1,j) and (i,j+1), using a visited set to avoid duplicates. Discuss time complexity O(k log k) and space O(k), and compare with binary search on sum value if needed.
Pro tip: Mention that you would handle duplicates carefully by using a visited set or by skipping duplicate sums, and discuss how to avoid generating the same pair multiple times. Also, proactively ask about edge cases like k=0 or empty arrays to show thoroughness.
Ask about input sizes, whether arrays can have duplicates, if k is guaranteed ≤ n*m, and if the output should be sorted by sum. Confirm the definition of 'k smallest sums' and whether ties should be included.
Propose a min-heap (priority queue) approach: initialize with (0,0), then repeatedly pop the smallest sum and push (i+1,j) and (i,j+1) if within bounds and not visited. Alternatively, discuss binary search on the sum value to find the k-th smallest sum and then collect all pairs ≤ that sum.
Use a visited set (e.g., a 2D boolean array or hash set of encoded indices) to avoid pushing the same pair multiple times. If duplicates in sums are allowed, ensure the algorithm still returns exactly k pairs (or all pairs if ties extend beyond k).
State that the heap approach takes O(k log k) time and O(k) space. Mention that binary search takes O((n+m) log(maxSum)) time but may require additional steps to collect all pairs. Discuss which is better based on k relative to n*m.
Walk through a small example (e.g., arr1=[1,7,11], arr2=[2,4,6], k=3) to demonstrate correctness. Test edge cases: k=0, k > n*m, empty arrays, negative numbers, and large k.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.