Remove Stones to Minimize the Total

class Solution {
public:
    int reduceStones(vector<int>& stones, int operations) {
        priority_queue<int> maxHeap(stones.begin(), stones.end());
        int currentSum = accumulate(stones.begin(), stones.end(), 0);
        
        for (int op = 0; op < operations; op++) {
            int topStone = maxHeap.top();
            maxHeap.pop();
            int reducedValue = topStone / 2;
            currentSum -= reducedValue;
            maxHeap.push(topStone - reducedValue);
        }
        
        return currentSum;
    }
};

In the C++ solution for minimizing the total number of stones through a series of operations, the approach utilizes a max-heap and operations to decrease the largest stone by half each time to reduce the overall sum of stones effectively.

• Create a max-heap from the given list of stones to always access the largest stone quickly.
• Calculate the initial sum of all stones.
• For a given number of operations, repeatedly remove the largest stone from the heap, reduce its value by half, and then place the reduced stone back into the heap.
• Adjust the total sum of stones by subtracting the amount reduced in each operation.
• Finally, return the new sum of stones after all operations have been performed.

This method ensures efficient manipulation and retrieval of stone quantities, crucial to minimizing the total number of stones as much as possible within the allowed operations.

class Solution {
    public int reducePiles(int[] stones, int maxRemoves) {
        PriorityQueue<Integer> maxHeap = new PriorityQueue<>((x, y) -> y - x);
        for (int stone: stones) {
            maxHeap.add(stone);
        }
        
        for (int i = 0; i < maxRemoves; i++) {
            int top = maxHeap.poll();
            maxHeap.add(top - (top / 2));
        }
        
        int total = 0;
        for (int stone: maxHeap) {
            total += stone;
        }
        
        return total;
    }
}

This Java program provides a solution to the problem of reducing the total number of stones by executing a specified number of removal operations. The goal is to minimize the total sum of stones left after a certain number of the largest piles are reduced by half a maximum number of times, defined by maxRemoves.

• Begin by initializing a PriorityQueue, maxHeap, which will sort elements in descending order. This helps efficiently access and remove the largest pile of stones.
• Populate the maxHeap with the array stones. Each element represents the count of stones in a pile.
• Iterate up to maxRemoves times, performing the following operations:
  1. Remove the largest pile from maxHeap.
  2. Calculate half of this pile and subtract it from the original pile size.
  3. Add the resultant value back to the maxHeap.
• After the operations, iterate over the elements of maxHeap and sum them up to compute the total amount of stones left.
• Return the total amount of stones.

This solution effectively uses the properties of the heap data structure, specifically a max-heap, to continuously identify and reduce the largest pile efficiently, thus minimizing the total number of stones in the given number of operations.

import heapq

class Solution:
    def reduceStones(self, stones: List[int], operations: int) -> int:
        max_heap = [-stone for stone in stones]
        heapq.heapify(max_heap)

        for _ in range(operations):
            largest_stone = -heapq.heappop(max_heap)
            reduced = largest_stone // 2
            heapq.heappush(max_heap, -(largest_stone - reduced))

        return -sum(max_heap)

The "Remove Stones to Minimize the Total" Python3 solution operates by transforming a list of stone weights into a max heap to efficiently perform operations aimed at minimizing the total weight. The implementation encompasses the use of Python's heapq module, which facilitates heap operations on a list interpreted as a heap.

• Initialize a list max_heap, converting each stone weight to a negative value to simulate a max heap using Python's min heap structure.
• Use heapq.heapify(max_heap) to transform the list into a heap.
• For each operation (up to the number specified):
  • Remove the largest element from the heap (-heapq.heappop(max_heap)), which simulates popping the maximum value.
  • Calculate the reduced weight by dividing the popped stone's weight by 2.
  • Insert the new weight back into the heap to preserve heap properties.
• Return the negated sum of all elements in the heap to provide the minimized total stone weight.

This approach ensures that the largest stones are reduced first, effectively decreasing the total weight as much as possible per operation. The use of a heap data structure optimizes the process of finding and removing the largest stone, ensuring that the solution efficiently handles operations within the constraint limits.