Minimize sum with halving operations

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.

What a Strong Answer Covers

• A clear algorithm with the right data structures and enough pseudocode or code-level detail to implement it.
• A correctness argument that explains why the algorithm covers all required cases.
• Time and space complexity, plus at least one alternative approach when relevant.
• Focused tests for normal cases, edge cases, and failure modes.

Follow-up Questions

• How would the approach change if the input were streaming or too large for memory?
• What invariants would you assert in production code?
• Which tests would catch off-by-one, duplicate, or tie-breaking bugs?

Minimize Sum with Halving Operations

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).

Hints

1. 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.
2. Maintain the elements in a max-heap so you can repeatedly extract the largest, halve it, and push it back in O(log n).
3. 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.

Solution

Solution Alignment

The prompt asks for an implementation-level answer. The safest way to present it is to define the state, maintain clear invariants, then walk through complexity and tests.

Problem Restatement

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.

Recommended Approach

Start with a brute-force baseline to confirm correctness, then identify the repeated work or ordering property that enables a better data structure such as a hash map, heap, stack, queue, two pointers, prefix sums, BFS/DFS, or dynamic programming. Write the implementation around a small invariant and test that invariant directly.

Correctness

The implementation should maintain an invariant after each loop or operation that directly matches the problem statement. At termination, that invariant implies the returned value has considered every valid candidate exactly once, or has preserved the required data-structure state after every API call.

Complexity

State the baseline complexity and the optimized complexity. For most interview constraints, justify why the optimized approach meets the expected input size.

Edge Cases and Tests

Empty and singleton inputs, duplicates, ties, invalid inputs, boundary values, and tests that exercise the main invariant.