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Problem 2530 Maximal Score After Applying K Operations

Problem Statement

Link - Problem 2530

Question

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

class Solution {
public:
    static long long maxKelements(vector<int>& nums, int k) {
        priority_queue<int> pq(nums.begin(), nums.end());
        long long score=0;
        for(int i=0; i<k; i++){
            int x=pq.top();
            score+=x;
            if (x==1){// early stop
                score+=(k-1-i);
                break;
            }
            pq.pop();
            pq.push((x+2)/3);
        }
        return score;
    }
};

Complexity Analysis

| Algorithm      | Time Complexity | Space Complexity |
| -------------- | --------------- | ---------------- |
| Priority Queue | O(n log n)      | O(n)             |

Explanation

Intial Thoughts

To approach this problem, we should first identify the key elements: the given array of integers, the number of operations, and the goal of maximizing the score. We should consider how the score changes after each operation and think about strategies to optimize it. Some initial ideas include prioritizing the largest numbers in the array, considering the impact of the ceiling function on the numbers, and exploring ways to maximize the overall score after multiple operations. We should also think about how to efficiently select the next number to operate on. Additionally, we should consider the constraints on the input values and the number of operations.

Intuitive Analysis

Intuitively, we want to focus on the largest numbers in the array first, as they will give us the most significant increase in score. Using a priority queue can help us keep track of the largest numbers efficiently. We should repeatedly select the largest number, add it to the score, and then update the number using the given formula. It’s essential to consider the point at which operating on a number no longer provides a significant increase in score, such as when the number becomes 1. At that point, we can use the remaining operations to add the minimum possible score. By following this greedy approach, we can intuitively solve the problem and find the maximum possible score.

1. Intuition

• To maximize the score, we need to choose the largest numbers from the array nums in each operation
• Using a priority queue pq allows us to efficiently select the maximum number in each operation
• The key insight is to replace the chosen number with its ceiling divided by 3, which is implemented as (x+2)/3
• This process is repeated k times to achieve the maximum possible score
• If the chosen number becomes 1, we can stop the process early because further operations will not increase the score
• The algorithmic choice of using a priority queue ensures that the solution works efficiently for large inputs
• The time complexity of the solution is O(k log n) due to the priority queue operations

2. Implementation

• The maxKelements function takes a vector nums and an integer k as input and returns the maximum possible score
• A priority queue pq is initialized with the elements of nums using pq(nums.begin(), nums.end())
• The score is initialized to 0 and incremented by the maximum number x in each operation using score += x
• The chosen number x is replaced with its ceiling divided by 3 using pq.push((x+2)/3)
• The process is repeated k times using a for loop for(int i=0; i<k; i++)
• If the chosen number becomes 1, the process is stopped early using if (x==1) and the remaining operations are accounted for using score+=(k-1-i)
• The final score is returned as a long long integer using return score

Complexity Analysis

Time Complexity:

• The time complexity of the given algorithm is dominated by two main operations: the construction of the priority_queue and the loop that pops and pushes elements from and to the queue. The construction of the priority_queue takes O(n log n) time because it involves heapifying the input array of size n.
• The loop that runs k times, where each iteration involves pq.top(), pq.pop(), and pq.push((x+2)/3), takes O(k log n) time in the worst case because each of these operations takes O(log n) time in a priority queue.
• Since the problem statement doesn’t provide any relation between n and k, we consider the worst-case scenario where k can be as large as n, resulting in a time complexity of O(n log n) due to the queue operations.

Space Complexity:

• The space complexity of the algorithm is primarily determined by the space required to store the input array and the priority_queue.
• The input array nums requires O(n) space because it contains n elements, and the priority_queue also requires O(n) space in the worst case because it can contain up to n elements.
• Therefore, the overall space complexity of the algorithm is O(n) due to the storage of the input array and the priority queue.

Footnote

This question is rated as Medium difficulty.

Hints

It is always optimal to select the greatest element in the array.

Use a heap to query for the maximum in O(log n) time.

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