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2558. Take Gifts From the Richest Pile (Easy)

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

Companies: DE Shaw

Related Topics:
Array, Heap (Priority Queue), Simulation

Similar Questions:

• Remove Stones to Minimize the Total (Medium)

Solution 1. Heap

// OJ: https://leetcode.com/problems/take-gifts-from-the-richest-pile
// Author: github.com/lzl124631x
// Time: O(KlogN)
// Space: O(N)
class Solution {
public:
    long long pickGifts(vector<int>& A, int k) {
        priority_queue<int> pq(begin(A), end(A));
        while (k-- && pq.size() && pq.top() > 1) {
            int n = pq.top(), s = sqrt(n);
            pq.pop();
            pq.push(s);
        }
        long long ans = 0;
        while (pq.size()) {
            ans += pq.top();
            pq.pop();
        }
        return ans;
    }
};