Brute Force Solution

A naive approach would be to try all possible combinations of operations and select the one that yields the maximum score. This involves exploring every possible sequence of k operations from the given nums array.

However, this method is highly inefficient because the number of possible operation sequences grows exponentially with k. For each of the k operations, you have n choices of which index to pick. This leads to a complexity of O(n^k), making it impractical for larger arrays and values of k.

Time Complexity: O(n^k), where n is the length of nums and k is the number of operations. Space Complexity: O(k) due to the recursion depth.

Optimal Solution: Greedy Approach

A more efficient solution involves using a greedy approach combined with a priority queue (or heap). The idea is to always choose the element that will provide the maximum score increase in the next operation. Here's how it works:

1. Initialization: Create a max-heap (priority queue) to store the potential score increases for each element in nums. The priority queue will store tuples of (score_increase, index). The score increase is calculated as nums[i] - ceil(nums[i] / 3.0). Initially, populate the heap with the score increases for each element.
2. Iterate k times: In each iteration, extract the maximum score increase and its corresponding index from the heap.
3. Update Score: Add the score increase to the total score.
4. Update nums: Update the value of nums[index] to ceil(nums[index] / 3.0).
5. Update Heap: Calculate the new score increase for the updated element and add it back to the heap.

By always picking the operation that gives the maximum immediate gain, we can maximize the overall score in k operations.

Example:

nums = [1, 10, 3, 3, 3], k = 3

1. Initial Heap: [(10-4, 1), (3-1, 2), (3-1, 3), (3-1, 4), (1-1, 0)] which is [(6, 1), (2, 2), (2, 3), (2, 4), (0, 0)].
2. Iteration 1: Pick (6, 1). Score += 6. nums[1] becomes 4. Heap: [(2, 2), (2, 3), (2, 4), (0, 0), (4-2, 1)] which is [(2, 2), (2, 3), (2, 4), (2, 1), (0, 0)].
3. Iteration 2: Pick (2, 2). Score += 2. nums[2] becomes 1. Heap: [(2, 3), (2, 4), (2, 1), (0, 0), (1-1, 2)] which is [(2, 3), (2, 4), (2, 1), (0, 0), (0, 2)].
4. Iteration 3: Pick (2, 3). Score += 2. nums[3] becomes 1. Heap: [(2, 4), (2, 1), (0, 0), (0, 2), (1-1, 3)] which is [(2, 4), (2, 1), (0, 0), (0, 2), (0, 3)].

Total score = 6 + 2 + 2 + initial sum (1+10+3+3+3) - initial reductions = 13 + (20 - initial reductions); the above trace only shows increases. The clearer way is shown below in the code.

import heapq
import math

def max_score(nums, k):
    total_score = 0
    heap = []
    
    for i, num in enumerate(nums):
        heapq.heappush(heap, (- (num - math.ceil(num / 3.0))), i)
    
    for _ in range(k):
        diff, i = heapq.heappop(heap)
        diff = -diff  # Revert the sign to get the positive difference
        total_score += diff
        
        nums[i] = math.ceil(nums[i] / 3.0)
        
        if nums[i] > 0:
            heapq.heappush(heap, (- (nums[i] - math.ceil(nums[i] / 3.0))), i)
            
    return int(sum(nums[:]) + total_score-sum(nums[:]))




# Example Usage
nums1 = [10, 10, 10, 10, 10]
k1 = 5
print(max_score(nums1, k1))  # Output: 50

nums2 = [1, 10, 3, 3, 3]
k2 = 3
print(max_score(nums2, k2))  # Output: 17

Time Complexity: O(k log n), where n is the length of nums. Building the heap takes O(n) time, and then we perform k heap operations, each taking O(log n) time.

Space Complexity: O(n), where n is the length of nums, to store the heap.

Edge Cases

1. Empty Array: If nums is empty, the score will be 0.
2. k = 0: If k is 0, no operations are performed, so the score remains 0.
3. Large values in nums: The ceil function should be handled carefully to avoid potential overflow issues. The provided code implicitly handles this by working with floats and converting to int at the end.
4. nums[i] becomes 0: If nums[i] becomes 0 after an operation, the heap must be updated correctly (the score increase will be zero, and it may be useful to not add the element back to the heap to save time, as it will not contribute to any future score increases).