Meta·Software Engineer·Onsite - Coding / Algorithms·Senior

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Apr 2026

Summary

Meta coding round, one problem the whole session. The question looked like a knapsack variant but the twist was you had to return the actual subset, not just the count. That detail changed everything about how you had to approach it.

Questions Asked (1)

Given a list of items each with a positive weight and a capacity W, find the maximum number of items whose total weight fits within W and return the actual subset of item IDs, not just the count.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

I went straight for DP and got maybe halfway through before realizing reconstructing the subset would be a mess.

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AI HintsAI Generated

Suggested Approach

Recognize that to maximize the number of items, you should always pick the lightest items first. Sort the items by weight ascending, then greedily add items until the next one would exceed the capacity. Return the IDs of the selected items.

Pro tip: Explicitly state that this greedy approach is optimal for maximizing count because any solution with more items would require replacing a heavier item with a lighter one, which is impossible after sorting. Also, mention that if the problem asked for maximum total weight, the approach would be different (e.g., 0/1 knapsack).

1. Clarify the problem

Confirm that the goal is to maximize the number of items, not the total weight, and that each item can be chosen at most once. Ask if the item IDs are unique and if the output should preserve any order.

2. Choose the right algorithm

Explain that a greedy strategy of selecting the lightest items first is optimal for maximizing count. Contrast with dynamic programming if the goal were to maximize weight.

3. Outline the steps

Sort the items by weight in ascending order. Iterate through the sorted list, adding items to the result as long as the cumulative weight does not exceed W. Stop when the next item would exceed W.

4. Analyze complexity

State that sorting takes O(n log n) time and the subsequent iteration takes O(n) time, so overall O(n log n) time. Space complexity is O(n) for the output (or O(1) extra if sorting in place).

5. Handle edge cases

Consider cases where no items fit (return empty list), all items fit (return all IDs), or multiple items have the same weight. Also discuss if the input is large and sorting might be optimized with counting sort if weights are bounded.

Key Points to Mention

Greedy algorithm: pick lightest items first to maximize count.
Proof of optimality: any solution with more items would require a lighter item set, which is impossible after sorting.
Time complexity: O(n log n) due to sorting, O(n) for selection.
Space complexity: O(n) for output, O(1) extra if in-place.
Edge cases: empty input, capacity too small, all items fit.
Contrast with 0/1 knapsack: if maximizing weight, use DP instead.

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