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2530. Maximal Score After Applying K Operations

Difficulty Topics

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

 

Constraints:

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

Solution

maximal-score-after-applying-k-operations.py
class Solution:
    def maxKelements(self, nums: List[int], k: int) -> int:
        heap = []
        res = 0

        for x in nums:
            heappush(heap, -x)

        for _ in range(k):
            if not heap: break

            x = -heappop(heap)

            res += x

            new = ceil(x / 3)
            if new > 0:
                heappush(heap, -new)

        return res