2530

2530. Maximal Score After Applying K Operations (Medium)

You are given a 0-indexed integer array nums and an integer k. You have a starting score of 0.

In one operation:

1. choose an index i such that 0 <= i < nums.length,
2. increase your score by nums[i], and
3. replace nums[i] with ceil(nums[i] / 3).

Return the maximum possible score you can attain after applying exactly k operations.

The ceiling function ceil(val) is the least integer greater than or equal to val.

 

Example 1:

Input: nums = [10,10,10,10,10], k = 5
Output: 50
Explanation: Apply the operation to each array element exactly once. The final score is 10 + 10 + 10 + 10 + 10 = 50.

Example 2:

Input: nums = [1,10,3,3,3], k = 3
Output: 17
Explanation: You can do the following operations:
Operation 1: Select i = 1, so nums becomes [1,4,3,3,3]. Your score increases by 10.
Operation 2: Select i = 1, so nums becomes [1,2,3,3,3]. Your score increases by 4.
Operation 3: Select i = 2, so nums becomes [1,1,1,3,3]. Your score increases by 3.
The final score is 10 + 4 + 3 = 17.

 

Constraints:

• 1 <= nums.length, k <= 105
• 1 <= nums[i] <= 109

Companies: McKinsey

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

Similar Questions:

• Sliding Window Maximum (Hard)
• Remove Stones to Minimize the Total (Medium)

Solution 1.

// OJ: https://leetcode.com/problems/maximal-score-after-applying-k-operations
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
    long long maxKelements(vector<int>& A, int k) {
        priority_queue<int> pq;
        for (int n : A) pq.push(n);
        long long ans = 0;
        while (k--) {
            int n = pq.top();
            pq.pop();
            ans += n;
            pq.push((n + 2) / 3);
        }
        return ans;
    }
};