Meta·Machine Learning Engineer·Technical Phone Screen·Senior

Senior

Jun 2026

Summary

Meta ML Engineer interview with a pretty gnarly algorithms problem focused on k-th smallest pair sum. No behavioral fluff, just straight into the deep end on heap-based approaches and complexity proofs.

Questions Asked (1)

Given two sorted non-decreasing arrays of non-negative integers and an integer k, return the k-th smallest value among all possible pair sums (one element from each array), without enumerating all pairs. Target O(k log min(n, m)) time and O(min(n, m)) space using a heap. Also discuss duplicate sums, edge cases like k=1 and k=n*m, and prove your approach is correct.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

This one hurt a little.

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AI HintsAI Generated

Suggested Approach

Use a min-heap to generate the smallest pair sums in order, starting with the smallest element from the shorter array paired with each element of the longer array. Pop the smallest sum, then push the next pair from the same row (incrementing the index in the longer array) until you have popped k sums. This yields O(k log min(n,m)) time and O(min(n,m)) space.

Pro tip: Clarify upfront that you assume both arrays are sorted non-decreasing and contain non-negative integers, and that k is 1-indexed. Mention that you would handle duplicates by allowing them in the heap and counting each popped sum as a distinct pair, unless the problem specifies distinct sums.

1. Clarify assumptions and edge cases

Confirm that arrays are sorted, non-negative, and that k is 1-indexed. Discuss edge cases: k=1 (minimum sum), k=n*m (maximum sum), empty arrays, and duplicate sums.

2. Choose the heap strategy

Select the shorter array (size m) to minimize heap size. Initialize a min-heap with pairs (A[i] + B[0], i, 0) for i=0..m-1, where A is the shorter array and B is the longer array.

3. Iteratively extract k smallest sums

Pop the smallest sum from the heap. If this is the k-th pop, return it. Otherwise, if the popped pair has index j < n-1 in B, push (A[i] + B[j+1], i, j+1) into the heap.

4. Prove correctness and analyze complexity

Argue that the heap always contains the next smallest candidate sums because each row is sorted. Time complexity: O(k log m) due to k pops and pushes. Space: O(m) for the heap.

5. Discuss duplicates and trade-offs

Explain how duplicates are handled (each pair is distinct). Compare with alternative approaches like binary search on value, noting that the heap method is optimal for small k and avoids enumerating all pairs.

Key Points to Mention

Use the shorter array to initialize the heap to achieve O(min(n,m)) space.
Each heap entry represents a row in the conceptual matrix of pair sums; advancing the column index generates the next sum in that row.
The heap invariant ensures the next smallest sum is always at the top because each row is sorted.
Time complexity O(k log min(n,m)) and space O(min(n,m)).
Edge cases: k=1 returns A[0]+B[0]; k=n*m returns A[-1]+B[-1]; handle empty arrays by returning None or raising an error.
Duplicates are counted as separate pairs unless the problem specifies distinct sums; the algorithm naturally includes them.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.