2530. Maximal Score After Applying K Operations

Problem Link

Description

---

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 [1,2,1,3,3]. Your score increases by 3.
The final score is 10 + 4 + 3 = 17.

 

Constraints:

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

Solution

---

Python3

class Solution:
    def maxKelements(self, nums: List[int], k: int) -> int:
        pq = []
        res = 0
 
        for x in nums:
            heappush(pq, -x)
        
        for _ in range(k):
            if not pq: break
 
            x = -heappop(pq)
            res += x
 
            nxt = ceil(x / 3)
            if nxt > 0:
                heappush(pq, -nxt)
        
        return res