MathWorks·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jul 2026

Summary

MathWorks software engineer interview with a classic task assignment optimization problem. Clean problem statement, but the efficiency constraint is what makes it interesting.

Questions Asked (1)

You have n tasks to split between two interns, where intern 1 gets exactly k tasks and intern 2 gets the rest. Each task has a different reward depending on which intern does it. How do you maximize the total reward? Your solution needs to handle n up to 10^5 efficiently.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

My first instinct was brute force and I immediately knew that wasn't going to fly at that scale.

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AI HintsAI Generated

Suggested Approach

Model the problem as selecting exactly k tasks for intern 1 to maximize total reward, where assigning a task to intern 1 yields a gain of (reward1 - reward2) over assigning it to intern 2. Compute these gains, sort them, and pick the k tasks with the largest positive gains. This greedy approach works because the choices are independent and the objective is linear.

Pro tip: Mention that this is equivalent to a maximum weight matching in a bipartite graph with a cardinality constraint, but the greedy solution is optimal due to the matroid structure. Also, discuss edge cases like k=0 or k=n and how to handle negative gains.

1. Understand the problem and define the objective

Restate the problem: assign exactly k tasks to intern 1 and n-k to intern 2 to maximize total reward. Define total reward as sum of rewards for each intern's assigned tasks.

2. Transform to a gain-based selection

For each task i, compute gain_i = reward1_i - reward2_i. Assigning task i to intern 1 instead of intern 2 increases total reward by gain_i. The baseline is assigning all tasks to intern 2.

3. Sort gains and select top k

Sort the gains in descending order. Select the k tasks with the largest gains to assign to intern 1. If some gains are negative, we still must pick exactly k, so we pick the k largest (which may include negatives).

4. Compute total reward and handle edge cases

Total reward = sum of all reward2_i + sum of selected gains. Handle edge cases: k=0 (all to intern 2), k=n (all to intern 1), and ties in gains (any selection works).

5. Analyze complexity and discuss alternatives

Sorting takes O(n log n) time, which is efficient for n=10^5. Mention that a heap-based selection could achieve O(n log k) but sorting is simpler. Also note that the greedy choice is optimal due to the independent gains.

Key Points to Mention

Greedy algorithm: select tasks with largest difference in rewards.
Proof of optimality: exchange argument or matroid intersection.
Time complexity: O(n log n) due to sorting, which is efficient for n=10^5.
Space complexity: O(n) for storing gains, or O(1) extra if sorting in place.
Handling negative gains: still must pick exactly k, so pick the k largest even if negative.
Edge cases: k=0, k=n, and when all gains are equal.

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