Oracle·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

May 2026

Summary

Oracle SWE interview with a greedy/heap problem that sounds straightforward but has a few layers to it. The question came with a lot of follow-ups baked in, which I wasn't fully expecting.

Questions Asked (1)

You have an array of non-negative integers and an integer k. Each operation lets you pick any element and replace it with the ceiling of half its value. Using at most k operations, minimize the total sum. Walk through an efficient algorithm, explain why it's correct, give the time and space complexity, and handle edge cases like zeros and very large k.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

I went straight to max-heap because you always want to halve the largest element first, and that part felt solid.

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AI HintsAI Generated

Suggested Approach

Use a max-heap to always apply the operation to the element that yields the largest reduction in sum. For each operation, pop the maximum, compute its halved value, add the reduction to the total sum, and push the new value back. Repeat until k operations are used or the maximum is 0.

Pro tip: Mention that the greedy choice is optimal because the reduction from halving is monotonic with the element's value, and using a heap ensures we always pick the best candidate. Also, note that if the maximum becomes 0, further operations are useless, so we can stop early.

1. Understand the problem and constraints

Clarify that each operation replaces an element x with ceil(x/2), and we want to minimize the sum using at most k operations. Note that zeros never change and large k may exceed useful operations.

2. Choose the right data structure

Use a max-heap to efficiently retrieve the largest element, as the reduction from halving is greatest for larger values. This allows O(log n) per operation.

3. Design the greedy algorithm

Initialize total sum. For each operation (up to k), pop the max, compute new value = ceil(max/2), update sum by subtracting (max - new value), and push new value if >0. Stop if max is 0.

4. Prove correctness

Argue that the greedy choice is optimal: at each step, halving the largest element gives the maximum possible reduction, and this local optimality leads to global optimality because reductions are independent and monotonic.

5. Analyze complexity and edge cases

Time: O((n + k) log n) for heap operations; Space: O(n). Handle zeros (skip), k=0 (return original sum), and very large k (stop when max is 0).

Key Points to Mention

Greedy strategy: always halve the current maximum element.
Max-heap implementation for efficient retrieval of the maximum.
Correctness proof: exchange argument or monotonicity of reduction.
Time complexity: O((n + k) log n) and space complexity: O(n).
Edge cases: zeros, k=0, k larger than needed (stop when max=0).
Optimization: if k is very large, we can stop early when all elements are 0.

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