Naive Approach

A brute-force approach would be to iterate through the array. If an element nums[i] is greater than 0, then iterate through all subarrays of size k starting at index i. If nums[i] is still greater than 0 after checking all such subarrays, or if we can't find a valid subarray, return false.

This approach would involve repeated iterations, resulting in a high time complexity.

Code (Python)

def solve_naive(nums, k):
    n = len(nums)

    def decrease_subarray(start_index, amount):
        for i in range(start_index, min(start_index + k, n)):
            nums[i] -= amount

    for i in range(n):
        if nums[i] > 0:
            can_decrease = False
            for j in range(min(i + 1, n - k + 1)):
              if i >= j and i < j + k:
                decrease_subarray(j, nums[i])
                can_decrease = True
                break
            if not can_decrease and nums[i] > 0:
              return False

    return all(x == 0 for x in nums)

Time Complexity: O(N*M), where N is the length of nums, and M is related to the number of subarrays that might need to be examined.

Space Complexity: O(1)

Optimal Approach

The optimal approach involves using a difference array to keep track of the changes made by the operations. Iterate through the nums array. At each index i, check if nums[i] is greater than the current accumulated decrease. If it is, then we need to apply the operation on the subarray starting at index i. Update the difference array to reflect this operation. If nums[i] is less than the current accumulated decrease, then it is impossible to make all elements zero.

Edge Cases

• k > len(nums): The subarray size cannot be greater than the array length; should return false.
• Empty array: Return true if the array is empty.
• All elements are zero initially: Return true.
• An element is negative during the process: return false

Code (Python)

def solve(nums, k):
    n = len(nums)
    diff = [0] * n
    curr_sum = 0

    for i in range(n):
        curr_sum += diff[i]
        if nums[i] < curr_sum:
            return False
        
        needed_reduction = nums[i] - curr_sum
        if needed_reduction > 0:
            if i + k > n:
                return False
            curr_sum += needed_reduction
            if i + k < n:
                diff[i + k] -= needed_reduction

    return True

Time Complexity: O(N), where N is the length of the nums array.

Space Complexity: O(N), due to the diff array.