Hello there! Are you ready to tackle another engaging problem today? We have a practical task that will enhance your problem-solving skills. This task involves working with arrays, leveraging Ruby hashes, sorting, and using techniques such as the two-pointer method that we've covered in previous Ruby lessons. Let's dive in!

Our task is as follows. Suppose you have two equally long arrays, a and b, with a length varying from 1 to 1000, with each element being a unique positive integer ranging from 1 up to 1,000,000. Your challenge is to craft a Ruby method that identifies the closest number in array b to 2 * b[i] for each i. Upon finding this number, say for a specific i, it is b[j], we need to construct a new array using a[j] values in the order of increasing i.

Consider this example:

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1a = [10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110]
2b = [4, 12, 3, 9, 6, 1, 5, 8, 37, 25, 100]

After executing your method, the resulting array should be:

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1[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. Double of this number is 8. The closest number to 8 in array b is 8 itself, located 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. Double of this number is 24. The closest number to 24 in b is 25, found at index 9. The corresponding value in a at this index is 100. Thus, 100 is added to our new array.

The third item in b is 3 at index 2. Double of this number is 6. The closest number to 6 in b is 6 itself, found at index 4. The corresponding value in a is 50. Therefore, 50 is added to our new array.

Proceed similarly for the remaining elements in b.

Begin by creating a sorted list from array b. This list will comprise pairs of values from b and their corresponding indices. Here, each pair's value represents the element in b, while the index denotes its location in array b.

This sorted list acts similarly to a hash map, organizing data for efficient retrieval and traversal. Here's the initial part of our Ruby method:

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1def find_and_replace(a, b)
2  b_sorted = b.map.with_index { |val, idx| [val, idx] }.sort_by { |pair| pair[0] }
3end

In this Ruby code, we utilize map.with_index to create an array of pairs. Each pair comprises a value from b and its index. The sort_by method then arranges these pairs based on their values in ascending order.

With our sorted list ready, we now initialize a right pointer j and a result array res. The pointer j helps maintain our search within the boundaries of b_sorted, while res holds our final output.

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1def find_and_replace(a, b)
2  b_sorted = b.map.with_index { |val, idx| [val, idx] }.sort_by { |pair| pair[0] }
3  
4  j = 0  # Initialize right pointer
5  res = Array.new(a.length, 0)  # Initialize the result array
6end

Here's where we implement the main logic. We iterate over each item in b_sorted using its index i. For every item, we calculate the target, which is twice the value of the current item in b_sorted. We adjust j until it points to the closest number smaller than the target (2 * b_sorted[i][0]). This continues while j is within bounds and the next number is closer to the target.

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1def find_and_replace(a, b)
2  b_sorted = b.map.with_index { |val, idx| [val, idx] }.sort_by { |pair| pair[0] }
3  
4  j = 0  # Initialize right pointer
5  res = Array.new(a.length, 0)  # Initialize the result array
6  
7  b.each_with_index do |val, i|
8    target = 2 * b_sorted[i][0]  # The target is twice the current number in the sorted B
9    while j < b_sorted.length - 1 && b_sorted[j + 1][0] < target
10      j += 1  # Move right pointer to find a number smaller than or equal to the target
11    end
12    if j < b_sorted.length - 1 && (b_sorted[j + 1][0] - target).abs < (target - b_sorted[j][0]).abs
13      j += 1  # Move pointer if next number is closer
14    end
15    res[b_sorted[i][1]] = a[b_sorted[j][1]]
16  end
17end

In this final step, we use the indices from b_sorted to modify the corresponding elements in array a. The right pointer j helps determine which element from a goes into res based on the same index relation.

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1def find_and_replace(a, b)
2  b_sorted = b.map.with_index { |val, idx| [val, idx] }.sort_by { |pair| pair[0] }
3  
4  j = 0  # Initialize right pointer
5  res = Array.new(a.length, 0)  # Initialize the result array
6  
7  b.each_with_index do |val, i|
8    target = 2 * b_sorted[i][0]  # The target is twice the current number in the sorted B
9    while j < b_sorted.length - 1 && b_sorted[j + 1][0] < target
10      j += 1  # Move right pointer to find a number smaller than or equal to the target
11    end
12    if j < b_sorted.length - 1 && (b_sorted[j + 1][0] - target).abs < (target - b_sorted[j][0]).abs
13      j += 1  # Move pointer if next number is closer
14    end    
15    res[b_sorted[i][1]] = a[b_sorted[j][1]]
16  end
17
18  res
19end

And there you have it! You've successfully crafted a Ruby method capable of producing the required output. This method iteratively replaces each element in array a according to the defined logic and returns the modified array.

It's crucial to comprehend the computational complexity of our Two-Pointer approach and its effectiveness for this problem.

1. Time Complexity: The key operations involve sorting the array b and traversing it with two pointers. Sorting an array of n elements in Ruby has a time complexity of O(n log n). The two-pointer traversal of the sorted array adds another O(n) complexity. Thus, the overall time complexity is dominated by the sorting operation, O(n log n).

2. Space Complexity: Aside from the input data, we require space for b_sorted and res, both having lengths identical to the input arrays. Hence, the space complexity is O(n), where n is the length of the input arrays.

Great job! You've tackled a sophisticated task that involved manipulating arrays and implementing advanced techniques such as sorting and the two-pointer method, all within Ruby. Through this exercise, you have honed your Ruby coding skills and demonstrated your ability to solve complex, real-world challenges.

Now it's your turn to master these techniques. We encourage you to practice similar challenges using what you've learned today. As you continue to practice, your Ruby problem-solving skills will only improve. Happy coding!