Welcome! Today, I am excited to guide you through a fascinating task involving lists in Kotlin: pairing up 'opposite' elements. Specifically, we're going to learn how to access and manipulate elements within a Kotlin List. This task provides an excellent opportunity to elevate your list-handling skills using the Kotlin language. Are you ready to get started? Let's dive right in!

Our task today is to form pairs of 'opposite' elements in a given List of integers. In a list consisting of n elements, the first and the last elements are considered 'opposite', the second element and the second-to-last element are considered 'opposite', and so forth. For a list with an odd length, the middle element is its own 'opposite'.

You will be provided with a List of n integers, where n could range from 1 to 100, inclusive. The task requires you to return a List of String objects. Each String consists of an element and its 'opposite' element joined by a space.

Let's use the example list numbers as listOf(1, 2, 3, 4, 5) to simplify our understanding. In this case, the output of our solution(numbers) function will be listOf("1 5", "2 4", "3 3", "4 2", "5 1").

Before we start writing code, let's familiarize ourselves with how to access elements of a list in Kotlin.

In Kotlin, the i-th element of a List numbers can be accessed as numbers[i], with the index starting from 0. Consequently, the first element is numbers[0], the second one is numbers[1], and so forth, up to numbers[numbers.size - 1] for the last element.

Kotlin
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1fun solution(numbers: List<Int>): List<String> {
2    val i = 0
3    val elementAtGivenIndex = numbers[i]
4    // Further code will follow
5}

Now, let's figure out how to access an element's 'opposite'.

The 'opposite' of the i-th element of the List is the element at the numbers.size - i - 1-th position. To illustrate this concept, consider standing at the start of a line and your friend standing at the end of the line. In this scenario, you and your friend could be considered 'opposites'. Similarly, the 'opposite' of numbers[0] is numbers[numbers.size - 0 - 1], the 'opposite' of numbers[1] is numbers[numbers.size - 1 - 1], and so forth.

Now that we understand how to locate an element's 'opposite', we can proceed to code our solution. Let's start by initializing an empty List named result to store our 'opposite' pairs and compute the list's size for future reference.

Kotlin
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1fun solution(numbers: List<Int>): List<String> {
2    val result = mutableListOf<String>()
3    val n = numbers.size
4    // Further code will follow
5}

The next step is to loop over all elements in our numbers list. Within our loop, we'll create a String for each pairing of 'opposite' elements. This String will consist of the i-th element and the n - i - 1-th element separated by a space. This pair will then be added to our result list. Here's the final version of our function:

Kotlin
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1fun solution(numbers: List<Int>): List<String> {
2    val result = mutableListOf<String>()
3    val n = numbers.size
4    for (i in numbers.indices) {
5        result.add("${numbers[i]} ${numbers[n - i - 1]}")
6    }
7    return result
8}
9
10fun main() {
11    val numbers = listOf(1, 2, 3, 4, 5)
12    println(solution(numbers))
13}

This function iterates over all the elements of the list. For each of these elements, it forms a String with its 'opposite' and subsequently adds the pair to the result list.

Great job! You've successfully navigated through the concept of 'opposite' pairs and list indexing in Kotlin. By now, you should be familiar with the notion of accessing and pairing elements in a list based on their positions. This fundamental step brings you closer to mastering list manipulation within Kotlin. Up next, we've prepared a series of hands-on exercises for you to apply and further practice what you've learned today. Remember, practice makes perfect. Keep up the great work, and happy coding!