Okay, let's analyze the problem of finding valid splits in an array.

Naive Solution

The most straightforward approach is to iterate through all possible split positions and, for each position, calculate the sum of the left and right subarrays. We then check if the condition sum(left) >= sum(right) is met. This approach is easy to understand but not the most efficient.

Code Example (Python):

def count_valid_splits_naive(nums):
    n = len(nums)
    count = 0
    for i in range(n - 1):
        left_sum = sum(nums[:i + 1])
        right_sum = sum(nums[i + 1:])
        if left_sum >= right_sum:
            count += 1
    return count

Big O Analysis:

• Time Complexity: O(n^2) because, for each split, we are summing subarrays which take O(n) time.
• Space Complexity: O(1) because we only use a constant amount of extra space.

Optimal Solution

To improve the time complexity, we can precompute the total sum of the array and then iterate through possible split positions, updating the left and right sums incrementally. This avoids repeatedly calculating sums.

Algorithm:

1. Calculate the total sum of the array.
2. Initialize left_sum to 0.
3. Iterate from i = 0 to n - 2:
   • Add nums[i] to left_sum.
   • Calculate right_sum = total_sum - left_sum.
   • If left_sum >= right_sum, increment the count.
4. Return the count.

Code Example (Python):

def count_valid_splits_optimal(nums):
    n = len(nums)
    total_sum = sum(nums)
    left_sum = 0
    count = 0
    for i in range(n - 1):
        left_sum += nums[i]
        right_sum = total_sum - left_sum
        if left_sum >= right_sum:
            count += 1
    return count

Big O Analysis:

• Time Complexity: O(n) because we iterate through the array once.
• Space Complexity: O(1) because we use a constant amount of extra space.

Edge Cases

• Empty Array: The problem statement specifies that the array has at least 2 elements, so we do not need to consider this case.
• All Negative Numbers: The algorithm works correctly even if all numbers are negative.
• All Zeroes: The algorithm works correctly even if all numbers are zero.

Summary

The optimal solution significantly improves the time complexity by precomputing the total sum and updating left and right sums incrementally. This reduces the time complexity from O(n^2) to O(n), making it much more efficient for large arrays.