• 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

Solution: Priority Queue

Keep all numbers in a priority queue (max heap), each time extract the top one (largest one), then put num – sqrt(num) back to the queue.

Tip: We can early return if all the numbers become 1.

Time complexity: O(n + klogn)
Space complexity: O(n)

// Author: Huahua
class Solution {
public:
  long long pickGifts(vector<int>& gifts, int k) {
    long long ans = accumulate(begin(gifts), end(gifts), 0LL);
    priority_queue<int> q(begin(gifts), end(gifts));
    while (k-- && q.top() > 1) {
      int cur = q.top(); q.pop();
      int next = sqrt(cur);
      ans -= (cur - next);
      q.push(next);
    }
    return ans;
  }
};

The post 花花酱 LeetCode 2558. Take Gifts From the Richest Pile appeared first on Huahua's Tech Road.