Take Gifts From the Richest Pile

Updated on 26 June, 2025
Take Gifts From the Richest Pile header image

Problem Statement

In this problem, you are provided with an array called gifts, where each element represents the number of gifts in different piles. Every second, you will perform several operations on these piles:

  • Identify the pile(s) with the highest number of gifts.
  • If multiple piles have the same maximum number of gifts, you may pick any of them.
  • For the chosen pile, compute the floor of the square root of its current number of gifts and update the pile with this value.

After executing these operations for k seconds, you are required to calculate and return the total number of gifts remaining across all piles.

Examples

Example 1

Input:

gifts = [25,64,9,4,100], k = 4

Output:

29

Explanation:

The gifts are taken in the following way:
- In the first second, the last pile is chosen and 10 gifts are left behind.
- Then the second pile is chosen and 8 gifts are left behind.
- After that the first pile is chosen and 5 gifts are left behind.
- Finally, the last pile is chosen again and 3 gifts are left behind.
The final remaining gifts are [5,8,9,4,3], so the total number of gifts remaining is 29.

Example 2

Input:

gifts = [1,1,1,1], k = 4

Output:

4

Explanation:

In this case, regardless which pile you choose, you have to leave behind 1 gift in each pile.
That is, you can't take any pile with you.
So, the total gifts remaining are 4.

Constraints

  • 1 <= gifts.length <= 103
  • 1 <= gifts[i] <= 109
  • 1 <= k <= 103

Approach and Intuition

Understanding the Operations

  1. Identifying the maximum:
    • At each second, the pile with the maximum number of gifts is targeted. This operation ensures that the reductions are impactful by focusing on the largest pile.
  2. Square root reduction:
    • The number of gifts in the selected pile is reduced to the floor of its square root. This specific operation dramatically reduces the pile size, especially for large numbers, enhancing the problem's complexity and strategy.

Steps in the Solution

  1. Parse through the array to find the pile with the maximum number of gifts.
  2. Apply the square root operation and update the pile.
  3. Repeat the above steps for k seconds.
  4. After k seconds, sum up the elements of the array to get the total number of gifts left.

Example Walkthrough

  • Example 1:

    • Initial piles: [25, 64, 9, 4, 100]
    • Operation sequence:
      • First second: Target 100 (max), replace with 10. New state: [25, 64, 9, 4, 10]
      • Second second: Target 64 (max), replace with 8. New state: [25, 8, 9, 4, 10]
      • Third second: Target 25 (max), replace with 5. New state: [5, 8, 9, 4, 10]
      • Fourth second: Target 10 (max), replace with 3. Final state: [5, 8, 9, 4, 3]
    • Total gifts remaining: 29
  • Example 2:

    • Given all elements are 1, no mathematical operation can reduce them further; hence, the total count remains unchanged despite the operations. Thus, the result is the sum of the elements, which is 4.

Through these examples and steps, we see how the operations uniquely manipulate the state of the piles, leading to the solution. This method ensures that the largest piles are drastically reduced, helping in reaching the minimal state at the end of k operations.

Solutions

  • C++
  • Java
  • Python
cpp
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.

java
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.

python
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.

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