DoorDash·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

May 2026

Summary

DoorDash software engineer interview with a greedy/sorting problem involving chefs and dishes, plus a follow-up array construction question. Pretty focused session, felt more like a technical phone screen than a full onsite loop.

Questions Asked (2)

You have three arrays: chef skill levels, dish difficulty levels, and dish profits. Each chef can cook at most one dish, but only if their skill meets or exceeds the dish's difficulty. Multiple chefs can independently cook the same dish and each earns the full profit. Return the maximum total profit you can get by assigning chefs to dishes.

Algorithms & Data Structures

Author's notes

Took me a second to realize multiple chefs can cook the same dish, which changes the whole approach.

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AI HintsAI Generated

Suggested Approach

Sort the chefs by skill and the dishes by difficulty, then use a max-heap to track the most profitable dishes that each chef can cook. Iterate through chefs in increasing skill order, adding all dishes they can cook to the heap, and assign the most profitable dish to each chef. Sum the profits of assigned dishes to get the maximum total profit.

Pro tip: Clarify that multiple chefs can cook the same dish and each earns the full profit, so the same dish can be selected multiple times. This means we don't need to remove dishes from the heap after assignment, and we can reuse the most profitable dish for all qualified chefs.

1. Understand the problem and constraints

Restate the problem: each chef can cook at most one dish if their skill >= dish difficulty, and multiple chefs can cook the same dish with full profit each. The goal is to maximize total profit.

2. Sort chefs and dishes

Sort the chefs by skill in ascending order and the dishes by difficulty in ascending order. This allows efficient matching of chefs to dishes they can cook.

3. Use a max-heap for profits

Iterate through chefs in increasing skill order. For each chef, add all dishes with difficulty <= chef's skill to a max-heap keyed by profit. Then, if the heap is not empty, assign the most profitable dish to the chef and add its profit to the total.

4. Sum and return total profit

After processing all chefs, return the accumulated total profit. The heap ensures each chef gets the highest possible profit dish they can cook.

Key Points to Mention

Sorting both arrays to enable efficient two-pointer or heap-based matching.
Using a max-heap (priority queue) to always pick the highest profit dish available for the current chef.
The greedy choice is optimal because each chef independently contributes the maximum possible profit without affecting others.
Time complexity: O(N log N + M log M + (N+M) log M) where N is number of chefs and M is number of dishes, dominated by sorting and heap operations.
Space complexity: O(M) for the heap in the worst case.
Edge cases: no chefs, no dishes, chefs with insufficient skill, or all dishes too difficult.

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

As a follow-up: given the dish difficulty and profit arrays, how would you build an array where each index represents a skill/difficulty level and stores the best profit a chef at that level could achieve?

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

This is basically the preprocessing step from the main problem pulled out on its own.

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AI HintsAI Generated

Suggested Approach

First, clarify that we need to precompute the maximum profit for each skill level by considering all dishes with difficulty ≤ that level. Then, sort dishes by difficulty, iterate through levels, and maintain the maximum profit seen so far, storing it in the result array.

Pro tip: Mention that this precomputation allows O(1) queries for any chef level, which is crucial for scalability in a real-time system like DoorDash. Also, note that if the input is large, we can avoid sorting by using a counting sort or bucket approach if difficulty range is small.

1. Clarify the problem

Confirm that the goal is to create an array where each index i represents skill level i and stores the maximum profit among all dishes with difficulty ≤ i.

2. Choose the right data structure

Decide to use an array of size maxDifficulty+1 (or maxSkill+1) to store the best profit for each level, initializing with 0 or -1 if no dish is available.

3. Process dishes efficiently

Sort the dishes by difficulty (or use counting sort if range is small) and iterate through them, updating the best profit for each difficulty level.

4. Propagate maximum profit

After processing all dishes, iterate through the array from left to right, ensuring each index stores the maximum profit seen so far (i.e., max of current and previous).

5. Handle edge cases and complexity

Discuss handling levels with no dishes (profit 0 or -1), and analyze time complexity: O(n log n) due to sorting, or O(n + D) with counting sort, where D is max difficulty.

Key Points to Mention

Precomputation for O(1) queries per chef level
Sorting dishes by difficulty or using counting sort for linear time
Maintaining a running maximum profit as we iterate through levels
Handling levels with no available dishes (e.g., profit 0 or -1)
Time and space complexity trade-offs (O(n log n) vs O(n + D))
Potential for further optimization if multiple queries are expected

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