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This question is commonly asked for Software Engineer candidates at Oracle during technical interviews.

Minimize sum with halving operations | Oracle Coding Question

Quick Overview

This interview question evaluates algorithm design, data structures, correctness, complexity, edge cases, and implementation details in a realistic interview setting. A strong answer for Minimize sum with halving operations states assumptions, handles edge cases, explains trade-offs, and shows how to validate the result clearly.

Minimize sum with halving operations

Company: Oracle

Role: Software Engineer

Category: Coding & Algorithms

Difficulty: Medium

Interview Round: Onsite

You are given an array of non-negative integers nums and an integer k. In one operation, choose any index i and replace nums[i] with ceil(nums[i] / 2). You may perform at most k operations. Return the minimum possible sum of the array after the operations. Describe an efficient algorithm, prove why it works, analyze time and space complexity, and discuss edge cases such as zeros and very large k.

Quick Answer: This interview question evaluates algorithm design, data structures, correctness, complexity, edge cases, and implementation details in a realistic interview setting. A strong answer for Minimize sum with halving operations states assumptions, handles edge cases, explains trade-offs, and shows how to validate the result clearly.

You are given an array of non-negative integers `nums` and an integer `k`. In one operation, you may choose any index `i` and replace `nums[i]` with `ceil(nums[i] / 2)` (integer division rounded up). You may perform **at most** `k` operations. Return the minimum possible sum of the array after performing the operations. The key insight is a greedy one: each operation removes value equal to `nums[i] - ceil(nums[i] / 2)` from the sum, and halving the current largest element always removes at least as much as halving any smaller element. So repeatedly halve the current maximum, which is maintained efficiently with a max-heap. Once the largest remaining element is `0`, no further operation can reduce the sum, so we stop early (this also handles the all-zeros case and very large `k`). **Examples** - `nums = [10, 20, 7]`, `k = 4` → `14`. Halve 20→10, 10→5, 10→5, 7→4, leaving `[5, 5, 4]`, sum 14. - `nums = [0, 0, 0]`, `k = 5` → `0`. Every element is already 0; no useful operation exists. - `nums = [1]`, `k = 100` → `1`. `ceil(1/2) = 1`, so halving 1 changes nothing; the loop stops when the max is reduced no further.

Constraints

1 <= nums.length <= 10^5 (the empty array is also handled and returns 0)
0 <= nums[i] <= 10^9
0 <= k <= 10^9 (k may far exceed the number of useful operations)
The sum can exceed 32-bit range; use a 64-bit accumulator (long / long long).

Examples

Input: ([10, 20, 7], 4)

Expected Output: 14

Explanation: Halve the current max each time: 20→10, 10→5, 10→5, 7→4, leaving [5,5,4] with sum 14.

Input: ([5, 19, 8, 1], 3)

Expected Output: 15

Explanation: 19→10, 10→5, 8→4 gives [5,5,4,1] = 15; spending ops on the largest values yields the biggest reductions.

Input: ([0, 0, 0], 5)

Expected Output: 0

Explanation: All elements are 0; the largest is 0, so the loop stops immediately and the sum stays 0.

Input: ([], 10)

Expected Output: 0

Explanation: Empty array — sum is 0 regardless of k.

Input: ([1], 100)

Expected Output: 1

Explanation: ceil(1/2)=1, so 1 can never shrink; once the max stays at 1 the reduction is 0 and the answer is 1 even with k=100.

Input: ([8], 3)

Expected Output: 1

Explanation: 8→4→2→1 using all 3 operations on the single element, ending at 1.

Input: ([3, 3, 3], 0)

Expected Output: 9

Explanation: k=0 means no operations are allowed, so the sum is unchanged at 3+3+3=9.

Input: ([100, 0, 50], 2)

Expected Output: 75

Explanation: Halve the two largest: 100→50, 50→25, leaving [25,0,50] = 75; the zero is never touched.

Hints

Each operation reduces the sum by nums[i] - ceil(nums[i] / 2). To minimize the final sum, greedily spend each operation where this reduction is largest — which is always the current maximum element.
Maintain the elements in a max-heap so you can repeatedly extract the largest, halve it, and push it back in O(log n).
Stop early when the largest remaining element is 0 (handles all-zeros input and very large k), and remember ceil(1/2) = 1 so a value of 1 cannot be reduced further.

Quick Overview