Welcome to this unit's coding session! We're going to take a deep dive into an exciting technique — the two-pointer technique. This technique is a crucial skill for enhancing your algorithmic problem-solving abilities, especially when dealing with arrays. In this unit, we'll apply this technique to a problem that involves an array of integers and a target value. Let's get started!

Envision the problem at hand: we've been given an array of distinct integers and a target value. The task is to find all pairs of integers from the given array that sum up to the target value using the two-pointer technique. The function findPairs should take this array of integers and a target value as parameters. It should return a MutableList containing Pair of numbers, sorted in ascending order by the first element of each pair. If no pairs satisfy this requirement, the function should return an empty MutableList.

Consider, for instance, an example in which you're given an array, val numbers = intArrayOf(1, 3, 5, 2, 8, -2), and target = 6. In this case, the function should return [( -2, 8 ), ( 1, 5 )] because only these pairs from the input array of integers add up to the target value.

The naive approach to solve this problem would be to use a pair of nested loops to check each pair of numbers.

This would have a time complexity of $O(n^2)$ and a space complexity of $O(n)$. The naive approach can be time-consuming and inefficient, particularly for large arrays.

Comparatively, the two-pointer technique makes the problem-solving process more efficient by eliminating unnecessary operations (repeatedly checking pairs that can't sum to the target), thus enhancing the overall performance and efficiency of the solution.

We first sort the array and then apply the two-pointer technique:

Two pointers, one at the start (left) and the other at the end (right), are initialized.

In each iteration, we calculate the sum of the two numbers accessed via these indices. If the sum is equal to the target value, the pair is added to the output list, and both pointers are moved toward the center. This is because we know there are no other potential pairs with the current values of numbers[left] and numbers[right] (since the numbers are distinct).

When the sum is less than the target, we increment the left index (increasing the value of numbers[left]), and when the sum is greater than the target, we decrement the right index (decreasing the value of numbers[right]). This process continues until the left index crosses the right index. At this point, all potential pairs have already been identified and added to the return list.