Greetings! Welcome to our lesson today, where we'll take a deep dive into an intriguing aspect of array manipulation. Have you ever considered the idea of traversing an array not from start to end or from end to start, but from the middle, extending towards both ends simultaneously? Today's lesson is all about this concept. Trust me, it's going to be fascinating! Let's jump right in.

Our task is as follows: Given an array of integers, we aim to return a new array that emerges from the center of the original array and alternates direction towards both ends. In other words, the first element of our new array will be the middle element of the original one. After establishing the starting point, we will alternate between the elements to the left and the right of the initial center until we have incorporated every element. If the length of the initial array is even, we first take the element to the left of the center, then the one to the right of the center, then do the alternation as described above.

For example, for numbers = [1, 2, 3, 4, 5], the output should be [3, 2, 4, 1, 5].

This task might initially seem complex; however, don't worry. We're going to break it down and construct our solution step-by-step. Keep in mind an additional condition: the length of the array — represented as n — can vary from 1 to 100,000, inclusive.

To start, we need to establish the middle point of our array. Why the middle, you ask? Well, our task necessitates that we traverse the array from the center to the ends. C#'s integer division operator, /, allows us to determine the middle. If the array has an odd length, we add the middle element to the newOrder list (as it isn't paired with any other element); otherwise, we keep it empty for now.

Here's a look at its application in the code:

C#
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1using System;
2using System.Collections.Generic;
3
4public class Program
5{
6    public static List<int> IterateMiddleToEnd(List<int> numbers)
7    {
8        int mid = numbers.Count / 2; // The index of the left middle element
9        List<int> newOrder;
10
11        if (numbers.Count % 2 == 1)
12        {
13            newOrder = new List<int> { numbers[mid] }; // Adding the middle element to the resulting array
14        }
15        else
16        {
17            newOrder = new List<int>(); // No elements in the resulting array for now
18        }
19
20        return newOrder;
21    }
22}

The resolution of our task requires two pointers: left and right. These pointers will originate from just before and just after the middle element, respectively.

Here's an illustration of how we can implement this in our C# function:

C#
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1using System;
2using System.Collections.Generic;
3
4public class Program
5{
6    public static List<int> IterateMiddleToEnd(List<int> numbers)
7    {
8        int mid = numbers.Count / 2; // The index of the left middle element
9        int left, right; // Pointer initialization
10        List<int> newOrder;
11
12        if (numbers.Count % 2 == 1)
13        {
14            left = mid - 1; // The left to the middle element
15            right = mid + 1; // The right to the middle element
16            newOrder = new List<int> { numbers[mid] }; // Adding the middle element to the resulting array
17        }
18        else
19        {
20            left = mid - 1; // Left middle element
21            right = mid; // Right middle element
22            newOrder = new List<int>(); // No elements in the resulting array for now
23        }
24
25        return newOrder;
26    }
27}

Now that our pointers are initialized, it's time to traverse the array and construct our new order. For this, we can effectively use a while loop in C#, which continues looping until the left pointer is less than zero, and the right pointer is not yet at the end of the list. In each iteration, we add the element at index left to the newOrder list, decrease the left pointer by one, add the element at index right, and increase the right pointer by one.

This is what the code looks like:

C#
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1using System;
2using System.Collections.Generic;
3
4public class Program
5{
6    public static List<int> IterateMiddleToEnd(List<int> numbers)
7    {
8        int mid = numbers.Count / 2; // The index of the left middle element
9        int left, right; // Pointer initialization
10        List<int> newOrder;
11
12        if (numbers.Count % 2 == 1)
13        {
14            left = mid - 1; // The left to the middle element
15            right = mid + 1; // The right to the middle element
16            newOrder = new List<int> { numbers[mid] }; // Adding the middle element to the resulting array
17        }
18        else
19        {
20            left = mid - 1; // Left middle element
21            right = mid; // Right middle element
22            newOrder = new List<int>(); // No elements in the resulting array for now
23        }
24
25        while (left >= 0 && right < numbers.Count)
26        {
27            newOrder.Add(numbers[left]);
28            newOrder.Add(numbers[right]);
29            left--;
30            right++;
31        }
32
33        return newOrder;
34    }
35
36    public static void Main(string[] args)
37    {
38        List<int> numbers = new List<int> { 1, 2, 3, 4, 5 };
39        List<int> result = IterateMiddleToEnd(numbers);
40
41        Console.WriteLine(string.Join(", ", result)); // Output: 3, 2, 4, 1, 5
42    }
43}

This way, we have created a new ordered list starting from the middle and alternating between left and right elements all the way to the end of the original array. This approach capably satisfies the requirements of our task!

Congratulations on reaching the end of this lesson! You've just uncovered a fascinating method of traversing and manipulating arrays! Be proud of yourself for comprehending this concept and implementing it in the code. As is always the case, practice makes perfect. Therefore, I encourage you to apply this concept to similar problems. Your journey in mastering C# and algorithms has just begun. Take the next step, and code away!