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Leetcode 2558. Take Gifts From the Richest Pile

Problem Statement

You are given an integer array gifts representing the number of gifts in several piles. Every seconds you select the richest pile and reduce the number of gifts from that pile to its integer square root. You are allowed to perform this operation k times in total. Return the total number of gifts remaining after k operations.

Example:

Example 1:
Input: gifts = [25,64,9,4,100], k = 4
Output: 29

Example 2:
Input: gifts = [1,1,1], k = 3
Output: 3

Clarifying Questions

  1. Can gifts contain zero or negative numbers?
    • No, gifts contains only positive integers.
  2. What should be done if two piles have the same maximum number of gifts?
    • You can select either pile with the maximum number of gifts.
  3. Will k always be less than or equal to the number of elements in gifts?
    • Not necessarily, k can be greater than the number of elements in gifts.

Strategy

  1. Implement a priority queue (max-heap) to efficiently get the richest pile.
  2. For each operation, perform the following steps:
    • Pop the maximum element from the heap.
    • Calculate its integer square root.
    • Push the square root back to the heap.
  3. Repeat this for k operations.
  4. Sum up the remaining elements in the heap to get the final result.

Code

#include <vector>
#include <queue>
#include <cmath>

using namespace std;

class Solution {
public:
    int totalGifts(vector<int>& gifts, int k) {
        // Use max heap (priority queue) to keep track of the richest piles
        priority_queue<int> heap(gifts.begin(), gifts.end());
        
        for (int i = 0; i < k; ++i) {
            // Extract the richest pile
            int richest = heap.top();
            heap.pop();
            
            // Reduce the number of gifts in the richest pile to its integer square root
            int reduced_gifts = static_cast<int>(sqrt(richest));
            
            // Push the reduced pile back into the heap
            heap.push(reduced_gifts);
        }
        
        // Calculate the total number of gifts remaining
        int total_sum = 0;
        while (!heap.empty()) {
            total_sum += heap.top();
            heap.pop();
        }
        
        return total_sum;
    }
};

Time Complexity

Therefore, the overall time complexity of this solution is O(n + k log n).

By following this strategy and using the provided solution, we can efficiently compute the total number of gifts remaining after performing the given operations.

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