This approach utilizes a two-pointer technique taking advantage of the sorted nature of the input array. The first pointer starts at the beginning of the array while the second starts at the end. By evaluating the sum at these two pointers, you can determine how to move the pointers:

• If the sum is greater than the target, move the right pointer left to decrease the sum.
• If the sum is less than the target, move the left pointer right to increase the sum.
• If the sum equals the target, you've found the solution.

This method operates in O(n) time and uses O(1) additional space.

This approach also takes advantage of the sorted array, integrating binary search for a more theoretically robust approach. For every element in the array, a binary search is employed to find the complement such that the sum is equal to the target:

• Iterate over the array with an index i.
• Compute the complement as target - numbers[i].
• Attempt to find this complement within remaining elements using binary search.
• The solution involves returning these two indices once found.
• Note: While this method theoretically can be less efficient, it remains simple to implement and understand.