Welcome! In this lesson, we're going to explore the performance efficiencies offered by utilizing Ruby's Hash. We will tackle an array-based problem requiring us to choose an optimal strategy to minimize the size of our array. Excited to dive in? Let's get started!

Our task is to manipulate an array of integers. You are required to construct a Ruby method titled minimal_max_block. This method should accept an array as input and examine an interesting property related to contiguous blocks within that array.

Specifically, you have to select a particular integer, k, from the array. Once you've selected k, the method should remove all occurrences of k from the array, resulting in multiple contiguous blocks, or sub-arrays. The unique feature of k is that it is chosen such that the maximum length among these blocks is minimized.

For example, consider the array [1, 2, 2, 3, 1, 4, 4, 4, 1, 2, 5]. If we eliminate all instances of 2 (our k), the remaining blocks would be [1], [3, 1, 4, 4, 4, 1], [5], with the longest containing 6 elements. Now, if we instead remove all instances of 1, the new remaining blocks would be [2, 2, 3], [4, 4, 4], [2, 5], with the longest containing 3 elements. Hence, the method should return 1 in this case, as it leads to a minimal maximum block length.

A straightforward way to address this problem is via a brute force method. Each value in the array can be tested by removing it and examining the sizes of the resulting sub-arrays.

Ruby
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1def minimal_max_block_bruteforce(arr)
2  min_max_block_size = Float::INFINITY
3  min_num = nil
4
5  arr.uniq.each do |num|  # Avoid duplicates.
6    indices = arr.each_index.select { |i| arr[i] == num }  # Indices where 'num' appears.
7    indices.unshift(-1).push(arr.length)  # Add artificial indices at the ends.
8    max_block_size = indices.each_cons(2).map { |x, y| y - x - 1 }.max  # Calculate max block size.
9
10    if max_block_size < min_max_block_size
11      min_max_block_size = max_block_size
12      min_num = num
13    end
14  end
15
16  min_num
17end

This approach is $O(n^2)$ in time complexity because it involves two nested loops: one iterating through each potential k value and another scanning the array for each of these k values. As n increases, this approach becomes impractical for large datasets, demonstrating a need for optimization.

Before exploring an optimized solution using Ruby Hash, let's comprehend why it is superior in terms of time and space complexity compared to the brute force approach.

1. Time Complexity: The optimized solution requires only a single traversal of the array, resulting in a linear time complexity, $O(n)$, where n is the number of elements. This is much more efficient than the $O(n^2)$ complexity of the brute force approach.

2. Space Complexity: The Hash-based solution utilizes two hashes to track the last occurrence and maximal block sizes for each number. In the worst-case scenario where all elements are unique, the space complexity is $O(n)$.

To find the number that, when removed, minimizes the length of the longest block, we need to track the last position of elements and the block size formed by each number. Ruby's Hash provides an efficient way to store this data.

First, we initialize two hashes. The last_occurrence hash stores the last occurrence index of each number, while max_block_sizes stores the maximum size of the blocks formed when each number is removed.

Ruby
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1def minimal_max_block(arr)
2  last_occurrence = {}
3  max_block_sizes = {}

We iterate over the array. For each number:

• If it's encountered for the first time, we consider the block from the array's start to its position, storing the size in max_block_sizes.
• If it has appeared before, calculate the block size and update max_block_sizes if necessary.
• Update last_occurrence with the current index.

Ruby
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1  arr.each_with_index do |num, i|
2    if last_occurrence[num].nil?
3      max_block_sizes[num] = i
4    else
5      block_size = i - last_occurrence[num] - 1
6      max_block_sizes[num] = [max_block_sizes[num], block_size].max
7    end
8    last_occurrence[num] = i
9  end

Calculate the "tail block" size, i.e., the block from the last occurrence of each number to the array's end, and update max_block_sizes if necessary.

Ruby
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1  last_occurrence.each do |num, pos|
2    block_size = arr.length - pos - 1
3    max_block_sizes[num] = [max_block_sizes[num], block_size].max
4  end

Identify the number with the smallest maximum block size in max_block_sizes and return it.

Ruby
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1  min_num = max_block_sizes.min_by { |_, size| size }.first
2
3  min_num
4end

Great job! In this lesson, we explored the advantages of using Ruby's Hash to optimize performance in array manipulations. We constructed a method to identify a specific number whose removal minimizes the maximum chunk length. By harnessing Ruby's Hash, we delivered a more efficient solution compared to the brute-force approach. Enjoy experimenting with these constructs further as you tackle interesting challenges in Ruby!