2558. Take Gifts From the Richest Pile

You are given an integer array gifts denoting the number of gifts in various piles. Every second, you do the following:

• Choose the pile with the maximum number of gifts.
• If there is more than one pile with the maximum number of gifts, choose any.
• Leave behind the floor of the square root of the number of gifts in the pile. Take the rest of the gifts.

Return the number of gifts remaining after k seconds.

 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

C++

• class Solution {
  public:
      long long pickGifts(vector<int>& gifts, int k) {
          make_heap(gifts.begin(), gifts.end());
          while (k--) {
              pop_heap(gifts.begin(), gifts.end());
              gifts.back() = sqrt(gifts.back());
              push_heap(gifts.begin(), gifts.end());
          }
          return accumulate(gifts.begin(), gifts.end(), 0LL);
      }
  };

JAVA

• class Solution {
      public long pickGifts(int[] gifts, int k) {
          PriorityQueue<Integer> pq = new PriorityQueue<>((a, b) -> b - a);
          for (int v : gifts) {
              pq.offer(v);
          }
          while (k-- > 0) {
              pq.offer((int) Math.sqrt(pq.poll()));
          }
          long ans = 0;
          for (int v : pq) {
              ans += v;
          }
          return ans;
      }
  }

PYTHON

• class Solution {
  public:
      long long pickGifts(vector<int>& gifts, int k) {
          make_heap(gifts.begin(), gifts.end());
          while (k--) {
              pop_heap(gifts.begin(), gifts.end());
              gifts.back() = sqrt(gifts.back());
              push_heap(gifts.begin(), gifts.end());
          }
          return accumulate(gifts.begin(), gifts.end(), 0LL);
      }
  };