#2530

Medium

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Maximal Score After Applying K Operations

Maximize your score by applying exactly k operations on an array using greedy selection and heap optimization techniques.

array greedy heap priority queue

Problem Statement

You are given a 0-indexed integer array nums and an integer k. Starting with a score of 0, you perform exactly k operations to increase your total score. In each operation, you select an index i, add nums[i] to your score, and update nums[i] to ceil(nums[i] / 3).

Return the maximum total score possible after performing exactly k operations. The problem tests greedy choice plus invariant validation by requiring selection of the current maximum value efficiently and updating it to maintain correctness for subsequent operations.

Examples

Example 1

Input: nums = [10,10,10,10,10], k = 5

Output: 50

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

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 Approach

Use a Max-Heap for Efficient Maximum Selection

Push all elements of nums into a max-heap. Each operation, pop the largest element, add it to the score, and push back the updated value ceil(value / 3). This ensures the greedy choice invariant is maintained at every step.

Track Score Accumulation

Initialize score to 0. For each of the k operations, extract the current maximum from the heap and increment the score. This direct accumulation avoids recomputation and guarantees the maximal total.

Maintain Heap Updates Correctly

After using a number, update it with ceil division and push it back into the heap. Failing to push the updated value back or incorrectly computing ceil can lead to suboptimal results.

Complexity Analysis

Metric	Value
Time	O(k \log n + n \log n)
Space	O(n)

Time complexity is O(k log n + n log n) due to heap operations and initial heap construction. Space complexity is O(n) for storing the heap elements.

What Interviewers Usually Probe

Ask about choosing a data structure to efficiently retrieve the largest element repeatedly.
Discuss the greedy invariant: always selecting the current maximum ensures optimal score.
Probe how ceil division affects subsequent operations and heap updates.

Common Pitfalls or Variants

Common pitfalls

Forgetting to push the updated value back into the heap after each operation.
Incorrectly computing ceil division which can reduce the total score.
Not handling large k efficiently, leading to timeouts without heap optimization.

Follow-up variants

Modify the division factor from 3 to any integer greater than 1.
Use a min-heap to find the minimal score instead of maximal by adjusting the greedy choice.
Allow choosing multiple elements in one operation and summing them before applying ceil division.

FAQ

What is the main strategy to maximize score in Maximal Score After Applying K Operations?

Always select the current maximum element for each operation using a max-heap and update it with ceil division to maintain the greedy invariant.

Why is a heap necessary for this problem?

A heap allows efficient retrieval and update of the maximum element in O(log n) time per operation, which is critical for large k and n.

Can we use a simple sort and array for selection?

Sorting each time is inefficient because it takes O(n log n) per operation, while a max-heap maintains the greedy selection in O(log n).

How does the ceil operation affect subsequent choices?

It reduces the chosen element for future operations, ensuring that greedy selection always considers the updated values to maintain correctness.

What are common mistakes when implementing this problem?

Not pushing the updated value back into the heap, miscalculating ceil division, or ignoring the heap structure for efficient max retrieval.

terminal

Solution

Solution 1: Priority Queue (Max Heap)

To maximize the sum of scores, we need to select the element with the maximum value at each step. Therefore, we can use a priority queue (max heap) to maintain the element with the maximum value.

1

2

3

4

5

6

7

8

9

10

class Solution:
    def maxKelements(self, nums: List[int], k: int) -> int:
        h = [-v for v in nums]
        heapify(h)
        ans = 0
        for _ in range(k):
            v = -heappop(h)
            ans += v
            heappush(h, -(ceil(v / 3)))
        return ans

Continue Practicing

#2542 maximum subsequence score #2551 put marbles in bags #2497 maximum star sum of a graph