Meta·Machine Learning Engineer·Technical Phone Screen·Senior

Senior

Jun 2026

Summary

Meta MLE coding round with a greedy array problem. Pretty clean problem once you see the trick, but I almost went with a brute force sort before the insight clicked.

Questions Asked (1)

Given two reward arrays of length n and an integer k, assign exactly k indices to mouse 1 (collecting reward1[i] for each) and the remaining n-k indices to mouse 2 (collecting reward2[i]). Return the maximum total reward.

Algorithms & Data Structures

Author's notes

The greedy insight took me a minute.

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AI HintsAI Generated

Suggested Approach

Start by assigning all indices to mouse 2 to get a baseline sum, then compute the gain (reward1[i] - reward2[i]) for each index. Sort these gains in descending order and add the top k gains to the baseline, which gives the maximum total reward.

Pro tip: Mention that this greedy approach is optimal because the gains are independent and sorting ensures we pick the k largest improvements. Also note the O(n log n) time complexity and O(n) space, and that it can be optimized to O(n) using a heap if k is small.

1. Understand the problem

Clarify that exactly k indices go to mouse 1 and the rest to mouse 2, and we want to maximize the sum of rewards. Restate the goal to ensure alignment.

2. Baseline assignment

Assume all indices are assigned to mouse 2 and compute the total reward as the sum of reward2. This serves as a reference point.

3. Compute gains

For each index i, calculate the gain if assigned to mouse 1 instead of mouse 2: gain[i] = reward1[i] - reward2[i].

4. Select top k gains

Sort the gains in descending order and pick the top k. Add these gains to the baseline sum to get the maximum total reward.

5. Analyze complexity and edge cases

Discuss time complexity O(n log n) due to sorting, and space O(n) for the gains array. Mention edge cases like k=0, k=n, and negative gains.

Key Points to Mention

Greedy strategy: selecting the k largest gains is optimal because each index's contribution is independent.
Mathematical transformation: total reward = sum(reward2) + sum of top k (reward1[i] - reward2[i]).
Sorting approach: O(n log n) time, O(n) space; can be optimized to O(n log k) using a min-heap if k is small.
Edge cases: k=0 (all mouse 2), k=n (all mouse 1), and negative gains (still must assign exactly k to mouse 1).
Proof of optimality: exchange argument or induction to show that any assignment not using the top k gains can be improved.
Potential follow-up: if rewards are large, use 64-bit integers to avoid overflow.

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