Amazon·Software Engineer·Online Assessment (OA)·Intermediate

Intermediate

Apr 2026

Summary

Amazon SWE online assessment with a greedy/heap problem that looks straightforward but has a nasty edge case around large m values. The math shortcut is the whole game here.

Questions Asked (1)

You have n VM types each with an initial inventory count. Process m rental requests one at a time: each request picks the VM type with the highest current inventory, earns revenue equal to that inventory value, then decrements it by 1. Compute total revenue across all m requests.

Algorithms & Data Structures

Author's notes

First instinct was just throw a max-heap at it and simulate.

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AI HintsAI Generated

Suggested Approach

Model the problem as repeatedly extracting the maximum inventory and then decrementing it. Use a max-heap (priority queue) to efficiently get the highest inventory in O(log n) per request, yielding O((n+m) log n) total time. For each of the m requests, pop the max, add it to revenue, decrement it, and if still positive, push it back.

Pro tip: Mention that if m is very large, you can optimize by processing in batches: when the top inventory is x and there are k types with that value, you can compute how many requests until the next change, reducing operations. This shows deeper algorithmic thinking.

1. Understand the problem and constraints

Clarify that each request selects the VM type with the highest current inventory, earns that inventory value, and then decrements it. Confirm that ties can be broken arbitrarily and that inventory can reach zero but not negative.

2. Choose the right data structure

Use a max-heap (priority queue) to efficiently retrieve and update the maximum inventory. This gives O(log n) per operation, which is optimal for this greedy simulation.

3. Simulate the process

For each of the m requests, extract the maximum inventory, add it to the total revenue, decrement it, and if the new value is greater than zero, reinsert it into the heap.

4. Analyze complexity and edge cases

The time complexity is O((n + m) log n) and space is O(n). Handle edge cases like all inventories zero (revenue 0) or m larger than total inventory (some requests earn 0).

5. Optimize if needed

If m is extremely large, consider batch processing or using a counting approach if inventory values are bounded. Discuss trade-offs between simplicity and performance.

Key Points to Mention

Greedy choice: always pick the current maximum inventory.
Max-heap (priority queue) for efficient maximum retrieval and updates.
Time complexity: O((n + m) log n) with heap; space O(n).
Handling ties: any VM type with the maximum inventory works.
Edge cases: zero inventories, m exceeding total inventory, large m.
Potential optimization: batch processing when many requests hit the same max value.

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