Pairing Opposite Elements in Kotlin Lists

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.

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

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.

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