How does the two-pointer approach solve the two-sum problem on a sorted array?

Place one pointer at the beginning and one at the end of the sorted array. Calculate the sum of elements at both pointers. If the sum equals the target, you found the pair. If the sum is too small, move the left pointer right to increase the sum. If the sum is too large, move the right pointer left to decrease the sum. Each step eliminates one position, so the algorithm runs in O of n time with O of one space. This works because the array is sorted: moving left increases and moving right decreases the sum monotonically. For the unsorted variant, use a hash map instead.