nums and store it in total.left and right, to 0.current to 0, which will track the sum of the current subarray.max_length to -1, which will track the maximum length of the subarray that sums to total - x.right pointer over the array, adding nums[right] to current.current is greater than total - x, subtract nums[left] from current and increment left.current equals total - x, update max_length with the maximum of max_length and right - left + 1.max_length is not -1, return the length of nums minus max_length as the minimum number of operations; otherwise, return -1.