How do you solve Maximal Score After Applying K Operations on LeetCode?

Maximal Score After Applying K Operations (LeetCode #2530) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(k log n) time complexity and O(n) space complexity. Key insight: Always choose the maximum element to maximize the score.

What is the brute force solution for Maximal Score After Applying K Operations?

The brute force approach for Maximal Score After Applying K Operations is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we will perform k operations by repeatedly selecting the maximum element from the array, adding it to our score, and then replacing it with its ceiling divided by 3. This is straightforward but inefficient as it requires scanning the array multiple times. The optimal Optimal Solution approach improves this to O(k log n).

What algorithm or data structure is used to solve Maximal Score After Applying K Operations?

Maximal Score After Applying K Operations uses the following concepts: Array, Greedy, Heap (Priority Queue). The recommended patterns to study are: Heap, Greedy.

What are the interview tips for Maximal Score After Applying K Operations?

Explain your thought process clearly when choosing data structures. Discuss time and space complexities during your solution presentation. Practice similar problems to recognize patterns and optimize your approach.

What are common mistakes when solving Maximal Score After Applying K Operations?

Not using a max-heap leading to inefficient solutions. Overlooking the ceiling function when updating values.

#2530

Maximal Score After Applying K Operations

Medium

ArrayGreedyHeap (Priority Queue)HeapGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(k log n)
Space	O(1)	O(n)

💡

Intuition

Time O(k log n)Space O(n)

In the optimal approach, we utilize a max-heap (priority queue) to efficiently get the maximum element in O(log n) time. This allows us to perform k operations in a much faster manner, significantly reducing the overall time complexity.

⚙️

Algorithm

3 steps

1Step 1: Create a max-heap from the nums array.
2Step 2: Initialize score to 0.
3Step 3: For k iterations, extract the maximum element from the heap, add it to the score, and push back its updated value (ceil(max_element / 3)).

solution.py11 lines

1import heapq
2
3def maxScore(nums, k):
4    max_heap = [-num for num in nums]  # Invert values for max-heap
5    heapq.heapify(max_heap)
6    score = 0
7    for _ in range(k):
8        max_val = -heapq.heappop(max_heap)
9        score += max_val
10        heapq.heappush(max_heap, -(-max_val // 3))  # ceil division
11    return score

ℹ

Complexity note: The time complexity is O(k log n) because each of the k operations involves extracting the maximum from the heap and then inserting the updated value back, both of which take O(log n) time. The space complexity is O(n) due to the storage of the heap.

1Always choose the maximum element to maximize the score.
2Using a heap allows efficient retrieval and updating of the maximum element.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.