Minimum Operations to Halve Array Sum

This problem asks for the minimum number of operations to reduce the sum of an array of positive integers by at least half. In each operation, you can choose any number and reduce it to exactly half its value.

Approach: Greedy with Max Heap

The most efficient approach leverages a greedy strategy combined with a max-heap data structure (priority queue). The core idea is this: to reduce the sum as quickly as possible, you should always halve the largest number in the array. This is because halving a larger number contributes more to the sum reduction than halving a smaller number.

1. Calculate the target reduction: First, compute half the sum of the input array nums. This is the target reduction we need to achieve.

2. Create a max-heap: A max-heap efficiently keeps track of the largest element. We populate the heap with all the numbers from nums.

3. Iterative halving: We repeatedly perform the following steps until the target reduction is met:

   • Extract the maximum element from the heap.
   • Halve the extracted element.
   • Subtract the reduction amount from the target reduction.
   • Insert the halved element back into the heap.
   • Increment the operation counter.

4. Return the operation count: The final count of operations represents the minimum number of operations required.

Code Implementation (Python)

import heapq
 
class Solution:
    def halveArray(self, nums: List[int]) -> int:
        total_sum = sum(nums)
        target_reduction = total_sum / 2
        heap = [-x for x in nums]  # Use negative values for max-heap
        heapq.heapify(heap)
        operations = 0
        current_reduction = 0
 
        while current_reduction < target_reduction:
            largest = -heapq.heappop(heap)
            halved_largest = largest / 2
            current_reduction += (largest - halved_largest)
            heapq.heappush(heap, -halved_largest)
            operations += 1
 
        return operations

Time and Space Complexity Analysis

• Time Complexity: The initial heap creation takes O(n) time. The while loop iterates at most O(n) times (because each operation at least halves the sum). Each heappop and heappush operation takes O(log n) time. Therefore, the overall time complexity is O(n log n).

• Space Complexity: The heap data structure uses O(n) space to store the array elements. The space for other variables is constant. Thus, the overall space complexity is O(n).

Other Languages (Conceptual Outline)

The core algorithm remains the same for other languages. The primary difference lies in the specific implementation of the priority queue (max-heap). Here's a conceptual outline for Java and C++:

Java: Use PriorityQueue<Double> with a custom comparator to create a max-heap.

C++: Use priority_queue<double> which is a max-heap by default.

Go: Go's container/heap package provides functionality to implement heaps. You'll need to define a custom heap type and implement the required methods. The logic remains the same as in Python.

The Python code above provides a complete and efficient solution. The other languages will follow similar steps using their respective heap implementations. Remember to adjust data types as necessary (e.g., using double for floating-point precision in halving operations).