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

Intermediate

Apr 2026

Summary

NVIDIA coding screen for a software engineer role, just one algorithmic problem on HackerRank. Pretty focused session, no fluff.

Questions Asked (1)

Given an array of positive integers and a number K, perform exactly K operations where each operation picks the largest element and replaces it with the ceiling of half its value. Return the minimum possible sum of the array after all K operations.

Algorithms & Data Structures

Author's notes

The greedy approach is pretty clear once you think about it: always shrink the biggest number first.

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AI HintsAI Generated

Suggested Approach

Use a max-heap to efficiently extract the largest element in each operation, replace it with its ceiling half, and reinsert. After K operations, sum the heap elements to get the minimum possible sum. This greedy strategy is optimal because reducing the largest element yields the greatest immediate decrease in sum.

Pro tip: Mention that for large K, the heap approach is O((N + K) log N), which is efficient. Also note that if K is very large, many elements become 1 and further operations don't change the sum, so you can early-exit when the max is 1.

1. Understand the problem and constraints

Clarify that we need exactly K operations, each on the current largest element, and we want to minimize the final sum. Consider edge cases like K=0, all elements 1, or very large K.

2. Choose the right data structure

Select a max-heap (priority queue) to efficiently retrieve and update the largest element in O(log N) time per operation.

3. Simulate the operations

For each of the K operations, pop the max, compute ceil(max/2), push it back, and optionally track the sum incrementally to avoid recomputing.

4. Compute and return the final sum

After K operations, sum all elements in the heap (or maintain a running sum) and return it as the minimum possible sum.

5. Analyze complexity and optimizations

State time complexity O((N + K) log N) and space O(N). Mention early termination if the maximum becomes 1, as further operations won't change the sum.

Key Points to Mention

Greedy choice: always reducing the current maximum minimizes the sum after each operation.
Max-heap (priority queue) for efficient extraction and insertion.
Ceiling of half: (x + 1) // 2 for integer arithmetic.
Time complexity: O((N + K) log N) and space O(N).
Early termination when the maximum element is 1, as further operations have no effect.
Handling large K: if K exceeds the number of operations needed to reduce all elements to 1, the sum stabilizes.

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