Welcome! Today, I am excited to guide you through an intriguing task involving arrays: pairing up 'opposite' elements. We're going to learn about accessing and manipulating elements in Scala arrays. This task provides an excellent opportunity to refine your array-handling skills in Scala. Are you ready to start? Let's dive right in!

Our task is to form pairs of 'opposite' elements in a given array of integers. In an array of n elements, we view the first and last elements as 'opposite', the second and second-to-last elements as 'opposite', and so on. If the array length is odd, the middle element is its own 'opposite'.

You are provided with an array of n integers, with n ranging from 1 to 100, inclusive. The task necessitates that you return an array of tuples, where each tuple comprises a pair of an element and its 'opposite' element.

For example, for numbers = Array(1, 2, 3, 4, 5, 6, 7), the output should be solution(numbers) = Array((1, 7), (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (7, 1)).

Before we start writing code, let's familiarize ourselves with how to access the elements of an array in Scala.

In Scala, the i-th element of an array numbers can be accessed as numbers(i), with the index starting from 0. Consequently, the first element can be accessed using numbers(0), the second using numbers(1), and so forth, up to numbers(numbers.length - 1) for the last element.

Already ready to embark? Brilliant! Now, let's figure out how to access each element's 'opposite'.

The 'opposite' of the i-th element of the array is the numbers.length - i - 1-th element. To visualize this, imagine that you are standing at the start of a line and your friend is at the end of the line, with both of you being 'opposites'. So, the opposite element for numbers(0) is numbers(numbers.length - 0 - 1), the opposite to numbers(1) is numbers(numbers.length - 1 - 1), and so on.

Now, let's start coding our solution. We initiate by initializing an empty list in which we will store our 'opposite' pairs and by calculating the array's length for later use.

Scala
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1def solution(numbers: Array[Int]): Array[(Int, Int)] = {
2  val result = scala.collection.mutable.ArrayBuffer[(Int, Int)]()
3  val n = numbers.length

Next, we loop over all elements in our array.

Scala
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1def solution(numbers: Array[Int]): Array[(Int, Int)] = {
2  val result = scala.collection.mutable.ArrayBuffer[(Int, Int)]()
3  val n = numbers.length
4  for (i <- 0 until n) {
5    // next step comes here
6  }

Within our loop, we create a tuple for each pair of 'opposite' elements. This pair includes the i-th element and the n - i - 1-th element. We then append this tuple to our result list. Here's the final version of our function:

Scala
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1def solution(numbers: Array[Int]): Array[(Int, Int)] = {
2  val result = scala.collection.mutable.ArrayBuffer[(Int, Int)]()
3  val n = numbers.length
4  for (i <- 0 until n) {
5    result.append((numbers(i), numbers(n - i - 1)))
6  }
7  result.toArray
8}

This function iterates over all elements of the array and, for each element, forms a pair with its 'opposite' and stores the pair as a tuple in the result list.

Brilliant work! You've just navigated through the concept of 'opposite' pairs and array indexing in Scala. By now, you should be comfortable with the idea of accessing and pairing elements in an array based on their positions. This is a fundamental step toward achieving proficiency in array manipulation in Scala. Up next, we've prepared some hands-on exercises for you to apply and practice what you've learned today. Remember, the more you practice, the better you improve. Keep going, and happy coding!