2530. Maximal Score After Applying K Operations LeetCode Solution

In this guide, you will get 2530. Maximal Score After Applying K Operations LeetCode Solution with the best time and space complexity. The solution to Maximal Score After Applying K Operations problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

  1. Problem Statement
  2. Complexity Analysis
  3. Maximal Score After Applying K Operations solution in C++
  4. Maximal Score After Applying K Operations solution in Java
  5. Maximal Score After Applying K Operations solution in Python
  6. Additional Resources
2530. Maximal Score After Applying K Operations LeetCode Solution image

Problem Statement of Maximal Score After Applying K Operations

You are given a 0-indexed integer array nums and an integer k. You have a starting score of 0.
In one operation:

choose an index i such that 0 <= i < nums.length,
increase your score by nums[i], and
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.

See also  2223. Sum of Scores of Built Strings LeetCode Solution

Constraints:

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

Complexity Analysis

  • Time Complexity: O(n\log k)
  • Space Complexity: O(n)

2530. Maximal Score After Applying K Operations LeetCode Solution in C++

class Solution {
 public:
  long long maxKelements(vector<int>& nums, int k) {
    long ans = 0;
    priority_queue<int> maxHeap;

    for (const int num : nums)
      maxHeap.push(num);

    for (int i = 0; i < k; ++i) {
      const int num = maxHeap.top();
      maxHeap.pop();
      ans += num;
      maxHeap.push((num + 2) / 3);
    }

    return ans;
  }
};
/* code provided by PROGIEZ */

2530. Maximal Score After Applying K Operations LeetCode Solution in Java

class Solution {
  public long maxKelements(int[] nums, int k) {
    long ans = 0;
    Queue<Integer> maxHeap = new PriorityQueue<>(Collections.reverseOrder());

    for (final int num : nums)
      maxHeap.offer(num);

    for (int i = 0; i < k; ++i) {
      final int num = maxHeap.poll();
      ans += num;
      maxHeap.offer((num + 2) / 3);
    }

    return ans;
  }
}
// code provided by PROGIEZ

2530. Maximal Score After Applying K Operations LeetCode Solution in Python

class Solution:
  def maxKelements(self, nums: list[int], k: int) -> int:
    ans = 0
    maxHeap = [-num for num in nums]
    heapq.heapify(maxHeap)

    for _ in range(k):
      num = -heapq.heappop(maxHeap)
      ans += num
      heapq.heappush(maxHeap, -math.ceil(num / 3))

    return ans
 # code by PROGIEZ

Additional Resources

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