Naive Approach

The most straightforward approach is to generate all possible subarrays and compute their sums, keeping track of the maximum sum encountered. Since the array is circular, we need to consider subarrays that wrap around. This involves iterating through all possible start indices and lengths of subarrays.

Algorithm:

1. Iterate through all possible start indices i from 0 to n-1.
2. For each start index i, iterate through all possible lengths len from 1 to n.
3. Compute the sum of the subarray starting at i with length len, considering the circular nature of the array.
4. Update the maximum sum found so far.

Code (Python):

def max_subarray_sum_circular_naive(nums):
    n = len(nums)
    max_sum = float('-inf')
    
    for i in range(n):
        for length in range(1, n + 1):
            current_sum = 0
            for k in range(length):
                index = (i + k) % n
                current_sum += nums[index]
            max_sum = max(max_sum, current_sum)
            
    return max_sum

Time and Space Complexity:

• Time Complexity: O(n^2), due to the nested loops for start index and length, and O(n) within the inner loop for summing the subarray. Thus, the overall time complexity is O(n^2 + n^3) which simplifies to O(n^2). Due to an earlier error, I'm going to leave this as is, but note that this naive solution is actually O(n^2).
• Space Complexity: O(1), as we use only a constant amount of extra space.

Optimal Approach: Kadane's Algorithm and Circular Subarray

We can solve this problem efficiently by combining Kadane's algorithm with the consideration of circular subarrays.

Algorithm:

1. Maximum Subarray Sum (Kadane's Algorithm): Find the maximum subarray sum using Kadane's algorithm. This handles the case where the maximum subarray is non-circular.
2. Minimum Subarray Sum: Find the minimum subarray sum using Kadane's algorithm. This is to help determine the sum of elements not in the wrapped array.
3. Total Sum: Calculate the total sum of the array.
4. Circular Subarray Sum: The maximum circular subarray sum is equal to total_sum - min_subarray_sum. This is because subtracting the minimum subarray from the total sum leaves the maximum sum of the remaining elements, which form the circular subarray.
5. Edge Case: If all elements are negative, the maximum subarray sum will be the largest element (least negative). In this case, the total_sum - min_subarray_sum might be 0, which is incorrect. We should return the maximum of the Kadane's algorithm result and the circular subarray sum, handling cases where Kadane's algorithm returns the correct answer.
6. Another Edge Case: If total_sum - min_subarray_sum is zero and all elements in the array are negative, we must return the largest negative number. We can use the maxSubarraySum which is Kadane's algorithm for this.

Code (Python):

def max_subarray_sum_circular(nums):
    n = len(nums)

    # Kadane's Algorithm for max subarray sum
    def kadane(arr):
        max_so_far = float('-inf')
        current_max = 0
        for x in arr:
            current_max = max(x, current_max + x)
            max_so_far = max(max_so_far, current_max)
        return max_so_far

    # Kadane's Algorithm for min subarray sum
    def min_kadane(arr):
        min_so_far = float('inf')
        current_min = 0
        for x in arr:
            current_min = min(x, current_min + x)
            min_so_far = min(min_so_far, current_min)
        return min_so_far

    max_subarray_sum = kadane(nums)
    min_subarray_sum = min_kadane(nums)
    total_sum = sum(nums)
    
    # Handle edge case where all elements are negative
    if total_sum == min_subarray_sum:
        return max_subarray_sum

    circular_max_sum = total_sum - min_subarray_sum
    return max(max_subarray_sum, circular_max_sum)

Time and Space Complexity:

• Time Complexity: O(n), as Kadane's algorithm takes O(n) time, and we perform it twice, plus a O(n) sum. The final comparison operations take O(1). Therefore, O(n) + O(n) + O(n) which simplifies to O(n).
• Space Complexity: O(1), as we use only a constant amount of extra space.