In this lesson, 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 slices. In this unit, we'll apply this technique to a problem that involves a slice of integers and a target value. Let's get started!

Envision the problem at hand: we've been given a slice of distinct integers and a target value. The task is to find all pairs of integers from the given slice that sum up to the target value using the two-pointer technique. The function FindPairs should take this slice of integers and a target value as parameters. It should return a slice containing pairs 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 slice.

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

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

This approach would have a time complexity of O(n^2) and a space complexity of O(1). The naive approach can be time-consuming and inefficient, particularly for large slices.

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 start by sorting the slice, preparing it for easier comparisons. We then apply the two-pointer technique:

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

2. In each iteration, we calculate the sum of the two numbers pointed to by the two respective pointers. If the sum is equal to the target value, the pair is added to the output slice, 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).

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

The first step to addressing this problem is sorting the slice. By sorting the slice of integers, we can use the two-pointer technique effectively, as we know that values on the left are always smaller, while those on the right are always larger. Let's start building our function with this step:

Go
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1package main
2
3import (
4    "sort"
5)
6
7func FindPairs(numbers []int, target int) [][]int {
8    // Sorting the slice
9    sort.Ints(numbers)
10    return [][]int{}
11}

Once the slice is sorted, let's move on to the next stage. We'll initialize two pointers — one at the start of the slice, left, and the other at the end of the slice, right. We'll also set up an empty slice of slices, pairs, to store our potential number pairs.

Go
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1func FindPairs(numbers []int, target int) [][]int {
2    sort.Ints(numbers)
3    left := 0
4    right := len(numbers) - 1
5    pairs := [][]int{}
6    return pairs
7}

The most challenging and exciting part is the formation of pairs. We'll run a for loop until the left pointer crosses the right one. In each iteration, if the total equals the target, we'll store the pair and move one step from both ends. If the total is smaller than the target, we'll move one step to the right from the left, seeking a larger value. Conversely, if the total is larger than the target, we'll move one step to the left from the right, seeking a smaller value. Here is how it could be done:

Go
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1func FindPairs(numbers []int, target int) [][]int {
2    sort.Ints(numbers)
3    left := 0
4    right := len(numbers) - 1
5    pairs := [][]int{}
6
7    // Using two-pointer technique to find pairs
8    for left < right {
9        total := numbers[left] + numbers[right]
10
11        if total == target {
12            // Add the pair to the result slice if the sum matches the target
13            pairs = append(pairs, []int{numbers[left], numbers[right]})
14            left++  // Move left pointer one step to the right
15            right-- // Move right pointer one step to the left
16        } else if total < target {
17            left++  // Move left pointer one step to the right to increase the sum
18        } else {
19            right-- // Move right pointer one step to the left to decrease the sum
20        }
21    }
22    return pairs // Return the slice of pairs that sum up to the target
23}
24
25// Example usage
26func main() {
27    numbers := []int{1, 3, 5, 2, 8, -2}
28    result := FindPairs(numbers, 6)
29    for _, pair := range result {
30        fmt.Printf("[%d, %d]\n", pair[0], pair[1])
31    }
32    // Output: [-2, 8]
33    //         [1, 5]
34}

In the FindPairs function, we first sort the slice, preparing it for easier pair identification. We then initialize two pointers and an empty slice to store the pairs. Using a for loop, we move the pointers according to the sum they point to, finding and storing pairs that add up to the target value. Finally, the pairs slice containing the valid pairs is returned.

• Sorting Operation: The sort.Ints(numbers) function sorts the slice in ascending order. The time complexity of this sorting step is O(n log n), where n is the number of elements in the slice.

• Two-Pointer Traversal: After sorting, the for loop traverses the slice with two pointers (left and right). Each pointer moves toward the center of the slice and makes a constant amount of work per iteration (either moving the left pointer, moving the right pointer, or both). The total number of iterations is linear in the number of elements, so the time complexity for this part is O(n). Since the sorting step dominates the time complexity, the overall time complexity of the function is O(n log n).

The space complexity is O(n) because, in the worst-case scenario (when all elements form a pair summing up to the target), the return slice would require space for n/2 pairs. The 1/2 is just a constant, and the polynomial expression is of order 1, hence O(n).

Congratulations on successfully handling a problem that requires the imperative two-pointer technique! This scenario involved understanding and meticulously applying the two-pointer technique to identify all relevant pairs of integers. In this task, we learned how summarizing concepts can ultimately aid us in finding a solution more efficiently, without having to approach each element explicitly.

Now, it's time for some practice! In the subsequent practice session, you'll handle similar problems that demand the use of the two-pointer method. Take everything you've learned from this lesson and apply it directly to the practice! Thank you for attending the session, and keep up the excellent work!