2530. Maximal Score After Applying K Operations

Difficulty: Medium

Topics: Array, Greedy, Heap (Priority Queue)

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

Hint:

1. It is always optimal to select the greatest element in the array.
2. Use a heap to query for the maximum in O(log n) time.

Solution:

Let's break down the solution for the "Maximal Score After Applying K Operations" problem:

Approach:

1. Use a Max Heap: The problem requires us to maximize the score by selecting the largest available number in nums for k operations. A Max Heap (priority queue) is ideal here since it allows us to efficiently extract the largest element and then insert updated values.

2. Extract the Maximum Element: For each operation, extract the largest element from the heap. This element will be added to the score.

3. Update the Extracted Element: After using the largest element, replace it with ceil(num / 3) (rounding up). This simulates the reduction in the value of the chosen element as per the problem's requirement.

4. Insert the Updated Element Back into the Heap: Insert the updated value back into the heap so that it can be used in subsequent operations if it remains among the largest values.

5. Repeat for k Operations: Perform this process exactly k times to ensure the maximum possible score is achieved after k operations.

Detailed Explanation:

• Initialization:

  • SplMaxHeap is used to maintain a max-heap (priority queue) to efficiently get the maximum element at each step.
  • ans keeps track of the cumulative score.
  • Insert all elements of nums into the max heap.

• Processing Each Operation:

  • Extract the largest element using extract().
  • Add this value to the sum.
  • Calculate the new value of this element as ceil(t / 3). Since PHP int division rounds down, we use (int)($t + 2) / 3 to simulate the ceiling.
  • Insert the updated value back into the heap.

• Return the Final Score: After k operations, return the accumulated sum.

Let's implement this solution in PHP: 2530. Maximal Score After Applying K Operations

<?php
/**
 * @param Integer[] $nums
 * @param Integer $k
 * @return Integer
 */
function maxKelements($nums, $k) {
    ...
    ...
    ...
    /**
     * go to ./solution.php
     */
}
?>

Example Walkthrough:

Example 1: nums = [10, 10, 10, 10, 10], k = 5

• Initial Heap: [10, 10, 10, 10, 10]
• Extract 10, score += 10, insert ceil(10/3) = 4: New Heap: [10, 10, 10, 10, 4]
• Extract 10, score += 10, insert ceil(10/3) = 4: New Heap: [10, 10, 4, 10, 4]
• Repeat similarly for all elements. The final score will be 50.

Example 2: nums = [1, 10, 3, 3, 3], k = 3

• Initial Heap: [10, 3, 3, 3, 1]
• Extract 10, score += 10, insert ceil(10/3) = 4: New Heap: [4, 3, 3, 3, 1]
• Extract 4, score += 4, insert ceil(4/3) = 2: New Heap: [3, 3, 3, 2, 1]
• Extract 3, score += 3, insert ceil(3/3) = 1: New Heap: [3, 3, 2, 1, 1]
• Final score is 17.

Complexity:

• Time Complexity: O(k log n), where k is the number of operations and n is the size of the heap. Each extraction and insertion operation on the heap takes O(log n) time, and this is done k times.
• Space Complexity: O(n) for storing the elements in the max heap.

This solution efficiently maximizes the score by always choosing the largest available number and managing the decreasing values using a priority queue.

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