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2530 - Maximal Score After Applying K Operations (Medium)

https://leetcode.com/problems/maximal-score-after-applying-k-operations/

Problem Statement​

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 <= 10^5
  • 1 <= nums[i] <= 10^9

Approach 1: Priority Queue​

Written by @wingkwong
class Solution {
public:
    long long maxKelements(vector<int>& nums, int k) {
        long long ans = 0;
        // we want to take the max one in each round
        priority_queue<int> pq(nums.begin(), nums.end());
        // perform k rounds
        while (k--) {
            // get the max one
            int t = pq.top(); 
            // pop it out
            pq.pop();
            // add to answer
            ans += t; 
            // add the ceil value
            // ceil(x / y) = (x + y - 1) / y
            // ceil(t / 3) = (t + 3 - 1) / 3 = (t + 2) / 3
            pq.push((t + 2) / 3);
        }
        return ans;
    }
};