Take Gifts From the Richest Pile

class Solution {
public:
    long long selectPresents(vector<int>& presents, int iterations) {
        // Use a max-heap for the presents
        priority_queue<int> maxHeap(presents.begin(), presents.end());
    
        // Process the heap 'iterations' times
        for (int j = 0; j < iterations; j++) {
            // Extract the largest present
            int largest = maxHeap.top();
            maxHeap.pop();
    
            // Reinsert modified largest value
            maxHeap.push(sqrt(largest));
        }
    
        // Sum all remaining items in the heap
        long long totalValue = 0;
        while (!maxHeap.empty()) {
            totalValue += maxHeap.top();
            maxHeap.pop();
        }
        return totalValue;
    }
};

The provided C++ code defines a function intended to select presents optimally from a pile using a strategy involving a max-heap data structure. Here's a concise breakdown:

• Starts by constructing a max-heap from an initial vector of present values, enabling efficient access to the largest present repeatedly.
• Iteratively modifies the heap for a given number of times (iterations). In each iteration, the largest present is removed from the heap, its square root is calculated (rounded down as an implicit part of integer operations), and then reinserted back into the heap.
• After modifying the heap through all iterations, the function computes the sum of all present values left in the heap to form the total value.

This procedure effectively makes the largest items smaller with each iteration, potentially optimizing the overall value extracted from the pile when the iteration strategy is carefully chosen. Additionally, using a heap ensures that each modification operation is efficiently handled, suitable for scenarios with large datasets and numerous iterations. The final computed value (totalValue) is returned as the long-term accumulation of the remaining present values in the max-heap.

public class Solution {
    
    public long gatherPresents(int[] presents, int n) {
        // Convert array to List while inverting values for max simulation
        List<Integer> presentedList = new ArrayList<>();
        for (int present : presents) {
            presentedList.add(-present);
        }
    
        // Create a min heap for negative values
        PriorityQueue<Integer> presentHeap = new PriorityQueue<>(presentedList);
        // Iteratively process the heap 'n' times
        for (int j = 0; j < n; j++) {
            // Extract the top (which is largest), then revert sign
            int largest = -presentHeap.poll();
    
            // Push back the modified largest value back into heap
            presentHeap.offer(-(int) Math.sqrt(largest));
        }
    
        // Total up remaining in heap by reverting negation
        long totalGifts = 0;
        while (!presentHeap.isEmpty()) {
            totalGifts -= presentHeap.poll();
        }
    
        return totalGifts;
    }
}

The Java solution provided efficiently tackles the problem of gathering presents from the richest pile using priority queues and mathematical operations. Below is a comprehensive breakdown of the approach used in the implementation:

• Begin by converting an array of present values into a list, while also negating the values. This transformation helps in simulating the extraction of maximum values using a min heap.

• A PriorityQueue named presentHeap is initialized with the negated values from the list. Java's priority queue serves as a min-heap by default, whereby negating the values effectively utilises it for maximum value extraction.

• The main computational loop runs n times:

  1. Extract the top value from the heap, which corresponds to the largest present due to the negation trick. The negation is then reverted to restore the original value.
  2. Calculate the square root of this largest value, negate it, and add it back into the heap.

• After processing n heaps' extractions and modifications, the sum of all values remaining in the heap is calculated, reverting the negation to count the total gifts.

• Each step optimally employs the data structure's properties and mathematical operations to efficiently gather the maximum possible gifts, culminating in returning the total count of gifts accumulated. This process ensures an effective usage of resources while maintaining clarity and performance efficiency.

class Solution:
    def selectPresents(self, presents, n_times):
        # Use a min-heap for the inverted values of the 'presents' list
        presents_heap = [-present for present in presents]
        heapq.heapify(presents_heap)
            
        # Conduct the operation 'n_times' times
        for _ in range(n_times):
            # Extract the largest element (in normal sense) in the heap
            top_val = -heapq.heappop(presents_heap)
                
            # Push the floor value of the square root of the top element back into the heap
            heapq.heappush(presents_heap, -math.floor(math.sqrt(top_val)))
            
        # Calculate the total sum of the values in the heap
        total_value = 0
        while presents_heap:
            total_value -= heapq.heappop(presents_heap)
            
        return total_value

The problem involves selecting gifts from piles, where each pile's value can be altered by taking the floor value of the square root of the largest pile, a specified number of times. This process of transformation and selection aims to maximize the total value of the gifts from the richest piles. The implementation utilizes Python3 with key data structure and algorithms detailed below.

• Begin by converting each value in the presents list to its negative. This transformation is necessary because Python's heapq module implements a min-heap, whereas the problem requires a max-heap behavior to always retrieve the largest present pile.
• Heapify the presents list to prepare it for the operations. This step organizes the data into a heap structure allowing efficient retrieval and updating of the roots.
• Use a loop to undergo the transformation n_times. In each iteration:
  1. Pop the maximum value (technically, the minimum negative value) from the heap.
  2. Calculate the floor of the square root of this value, negate it, and then push it back into the heap.
• After completing the transformations, calculate the total value by popping all elements from the heap and summing them up. As elements in the heap are negative, subtract each popped value from a total counter initialized at zero.
• Finally, return the computed total value.

This solution leverages the efficiency of the heap data structure, primarily in scenarios involving repeated access and modification of the largest element. With Python's heapq module providing the necessary operations of heapify, heappop, and heappush, the solution becomes both efficient and more straightforward to implement. This method ensures that even with large lists and a significantly high number of operations, the solution remains computationally feasible.