Maximal Score After Applying K Operations

Problem Statement

In this problem, you are presented with an array of integers, nums, each of which represents the points you can gain. With these points, you start with a score of 0. The objective is to maximize this score using exactly k operations. During each operation, you can:

1. Pick an index i within the bounds of the array (0 <= i < nums.length).
2. Add the value at this index (nums[i]) to your current score.
3. Modify the value at this index by applying the ceiling function to nums[i] / 3. This function rounds up the result of nums[i] / 3 to the nearest integer.

Your goal is to return the highest possible score after performing exactly k operations on the array.

Examples

Example 1

Input:

nums = [1, 10, 3], k = 3

Output:

17

Explanation:

Pick 10 → score = 10, nums = [1, 4, 3]  
Pick 4 → score = 14, nums = [1, 2, 3]  
Pick 3 → score = 17, nums = [1, 2, 1]

Example 2

Input:

nums = [5, 5, 5], k = 2

Output:

10

Explanation:

Pick any 5 → score = 5, value becomes 2  
Pick another 5 → score = 10

Constraints

• 1 <= nums.length, k <= 10^5
• 1 <= nums[i] <= 10^9

Approach and Intuition

Given the nature of the problem, a strategic approach can maximize the score:

1. Maximal Value Selection:

   • At each operation, aim to add the maximum value available in nums to maximize the score increment per operation.
   • This implies that we should prioritize indices with larger values due to the setup of the problem, as the bigger the number, the higher the score we can obtain from that operation.

2. Understanding Ceiling Division:

   • Post every operation, nums[i] is replaced with ceil(nums[i] / 3), meaning that each value can still contribute to the score in subsequent operations, albeit less significantly each time.
   • Since the smaller the number, the lesser it reduces when divided by three and then rounded up, selecting larger numbers earlier maintains a higher minimum value of each element over successive operations.

3. Iterative Picking with a Focus on Large Values:

   • If k is smaller than or equal to the length of nums, each element of nums might only be used once for maximization. However, if k is larger, some elements, particularly those with higher initial values, will need to be picked multiple times.
   • A priority system (like a max heap) can be effective here. We can keep picking the largest element from nums, update the score, modify the number as per the given rule, and then re-evaluate its standing in priority for future operations.

4. Edge Considerations:

   • If an element in nums is one of the few sizable elements in a mostly smaller array, reusing this element multiple times before it diminishes considerably becomes a plausible strategy.
   • By contrast, in a uniformly large array, the spread of operations might be more even.

Through these strategic selections and operations, the goal of achieving the maximum possible score can be approached systematically. Given that the problem constraints are quite significant (1 <= nums.length, k <= 10^5 and 1 <= nums[i] <= 10^9), an efficient implementation of the above strategy will be crucial for handling larger inputs optimally.

Solutions

class Solution:
    def maximumKElements(self, data: List[int], count: int) -> int:
        negative_heap = [-n for n in sorted(data, reverse=True)[:count]]
        result = 0
        
        for _ in range(count):
            top_element = heapq.heappop(negative_heap)
            result -= top_element
            heapq.heappush(negative_heap, top_element // 3)
        
        return result