Hello there! Are you ready to solve another engaging problem today? We have a practical task that will enhance your problem-solving skills. It involves critical aspects of programming — dealing with arrays and using techniques such as sorting and the two-pointer method. So, let's jump in!

Our task is as follows. Suppose you have two equally long arrays, A and B, with lengths varying from 1 to 1000, with each element being a unique positive integer ranging from $1$ up to $10^6$. Your challenge is to create a C# method that performs the following steps:

1. For each element B[i] in array B, double its value to get 2 * B[i].
2. Find the closest number to 2 * B[i] in array B. Let's call this closest number B[j].
3. For each index i in array B, get the value at index j in array A, i.e., A[j].
4. Create a new array where each element is A[j] corresponding to the closest number found in B.

To illustrate this, let's consider an example. We have:

C#
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1int[] A = { 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110 };
2int[] B = { 4, 12, 3, 9, 6, 1, 5, 8, 37, 25, 100 };

After running your method, the resulting array should look like this:

C#
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1int[] result = { 80, 100, 50, 20, 20, 60, 40, 20, 110, 90, 110 };

Let's walk through the first few steps:

The first item in B is 4 at index=0. Doubling this number gives us 8. The closest number to 8 in array B is 8, which is at index=7. The number at the same index in array A is 80, so we add 80 to our new array.

The second item in B is 12 at index=1. Doubling this number gives us 24. The closest number to 24 in B is 25, which is at index=9. The corresponding index in A has the number 100. So, we add 100 to our new array.

The third item in B is 3 at index=2. Doubling this number gives us 6. The closest number to 6 in B is 6, which is at index=4. The corresponding index in A has the number 50. So, we add 50 to our new array.

We continue this process for the rest of the elements in B.

Let's embark on our solution-building journey by constructing a sorted list for array B. This list will include pairs of values (val) and their corresponding indices (idx) from array B. Here, val represents the element in B, while idx denotes the index at which val is found in array B.

This sorted list will be similar to an associative array, storing 'value-index' pairs. It not only organizes the data for efficient retrieval but also makes it easier for us to traverse the list. Here's the introductory part of our C# method, including the creation of the sorted list:

C#
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1using System;
2using System.Collections.Generic;
3
4public class Solution
5{
6    public static int[] FindAndReplace(int[] A, int[] B)
7    {
8        List<(int value, int index)> B_sorted = new List<(int, int)>();
9        for (int i = 0; i < B.Length; i++)
10        {
11            B_sorted.Add((B[i], i));
12        }
13        B_sorted.Sort((a, b) => a.value.CompareTo(b.value));

In the above code, we generate a list of tuples comprising the values from B and their respective indices using a simple loop. Then, the Sort method arranges these tuples in ascending order of their values.

Now that our sorted list is ready, we initiate the right pointer, j, and the result array, res. The former assists in ensuring our search remains within the boundaries of B_sorted, while the latter will be our final output containing the replaced elements from array A. We'll update res as we progress through the solution.

C#
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1    int j = 0; // Initialize right pointer
2    int[] res = new int[A.Length]; // Initialize the result array

The primary logic of the problem lies in this step. We iterate over each item in B_sorted using its index i. For every item, we calculate the target, which is double the value of the element at the current index in B_sorted. We adjust the position of the right pointer, j, until it points to the closest number that is less than the target (2 * B_sorted[i].value). The operation continues as long as j is within the bound for B_sorted and the next number in the array is smaller than the target.

Once we've identified a number greater than or equal to the target, we compare it with the previous number to see which one is closer to the target. If the next number is closer, we advance j by one step.

C#
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1        for (int i = 0; i < B.Length; i++)
2        {
3            int target = 2 * B_sorted[i].value; // The target is twice the current number in the sorted B
4            while (j < B.Length - 1 && B_sorted[j + 1].value < target)
5            {
6                j++; // Move the right pointer to find a number smaller than or equal to the target
7            }
8            if (j < B.Length - 1 &&
9                Math.Abs(B_sorted[j + 1].value - target) < Math.Abs(target - B_sorted[j].value))
10            {
11                j++; // Move the right pointer one more step if the next number is closer to the target
12            }

In this final step, we employ the indices from B_sorted to alter the appropriate elements in array A. Based on the position of the right pointer j, we replace the corresponding element in res with the element in A located at the same index.

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1            res[B_sorted[i].index] = A[B_sorted[j].index];  
2            // Collect the corresponding element from A at the same index as the closest number in B_sorted
3        }
4
5        return res;
6    }
7}

The fully combined code for this solution is found below. It effectively finds and replaces elements in the array A based on the closest doubled values in the array B:

C#
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1using System;
2using System.Collections.Generic;
3
4public class Solution
5{
6    public static int[] FindAndReplace(int[] A, int[] B)
7    {
8        List<(int value, int index)> B_sorted = new List<(int, int)>();
9        for (int i = 0; i < B.Length; i++)
10        {
11            B_sorted.Add((B[i], i));
12        }
13        B_sorted.Sort((a, b) => a.value.CompareTo(b.value));
14
15        int j = 0; // Initialize right pointer
16        int[] res = new int[A.Length]; // Initialize the result array
17
18        for (int i = 0; i < B.Length; i++)
19        {
20            int target = 2 * B_sorted[i].value; // The target is twice the current number in the sorted B
21            while (j < B.Length - 1 && B_sorted[j + 1].value < target)
22            {
23                j++; // Move the right pointer to find a number smaller than or equal to the target
24            }
25            if (j < B.Length - 1 &&
26                Math.Abs(B_sorted[j + 1].value - target) < Math.Abs(target - B_sorted[j].value))
27            {
28                j++; // Move the right pointer one more step if the next number is closer to the target
29            }
30            res[B_sorted[i].index] = A[B_sorted[j].index];  
31            // Collect the corresponding element from A at the same index as the closest number in B_sorted
32        }
33
34        return res;
35    }
36
37    public static void Main()
38    {
39        int[] A = { 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110 };
40        int[] B = { 4, 12, 3, 9, 6, 1, 5, 8, 37, 25, 100 };
41        int[] result = FindAndReplace(A, B);
42        Console.WriteLine(string.Join(", ", result)); // 80, 100, 50, 20, 20, 60, 40, 20, 110, 90, 110
43    }
44}

It's vital to have an understanding of the computational complexity of our Two-Pointer approach and why it's effective for this problem.

1. Time Complexity: The main steps of our solution involve sorting the array B and traversing it with two pointers. Sorting an array of n elements has a time complexity of O(n log n). The two-pointer traversal of the sorted array adds an O(n) time complexity. Thus, the overall time complexity of our solution is O(n log n) for the sorting operation, which dominates the linear time traversal.

2. Space Complexity: Apart from the input data, our solution needs space for B_sorted and res, both of which are arrays having the same length as the input. Therefore, our solution has a linear space complexity of O(n), where n is the length of the input arrays.

Great job! You've successfully tackled a high-level task that involved manipulating arrays and implementing advanced techniques such as sorting and the two-pointer method. Through this exercise, you've honed your C# coding skills further and demonstrated your ability to solve a complex, real-world challenge.

Now it's your turn to master these techniques. We encourage you to practice solving similar challenges using the skills you've learned from this lesson. The more you practice, the better your problem-solving skills will become. So, get started and enjoy your journey in coding!