Hello and welcome to our exciting exploration of strings! For today's lesson, we've prepared something extraordinarily interesting: you will learn how to access characters from a string following a distinctive pattern. Our presentation is both comprehensive and concise, allowing you to master the concept promptly. Let's get started!

Imagine this: You receive a string from which you need to extract characters, but the sequence in which you select them diverges from the norm. You start with the first character, then select the last character, move to the second character, choose the second-to-last character, and continue this pattern until there are no characters left. Quite a mind-bender, isn't it?

Here's what we mean:

You are required to craft a Kotlin function, fun solution(inputString: String): String. This function takes inputString as a parameter, a string of lowercase English alphabet letters ('a' to 'z'), with a length ranging between 1 to 100 characters. The function then returns a new string, fashioned from the input string but with characters selected in the pattern we described above.

For example, if the inputString is "abcdefg", the function should return "agbfced".

Before we delve into the problem-solving aspect, let's arrange our result store. We initiate a variable, result, as an empty String to stockpile the output.

Kotlin
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1fun solution(inputString: String): String {
2    var result = ""

Upon initialization, we need to traverse the inputString. Kotlin furnishes a for loop to iterate over all elements in a string efficiently.

Now, the question arises: What is the requisite number of iterations for our loop? Given that we select two characters per iteration — one from the beginning and one from the end — we need the loop to run for half of the string's length if the length is even, or half the length plus one if it's odd to include the middle character.

We achieve this by employing 0 until length / 2 + length % 2. This ensures that the loop iterates for half the length if it is even and half the length plus one if it's odd.

Here's our function thus far:

Kotlin
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1fun solution(inputString: String): String {
2    var result = ""
3    val length = inputString.length
4    for (i in 0 until length / 2 + length % 2) {
5        // Implementation in next step
6    }
7}

Now that we are nested inside our loop, we can select characters and add them to our result.

First, we fetch the character from the beginning of the input string by calling inputString[i], and append this to our result.

Next, we secure the character from the end of the input string. We compute its index using length - 1 - i. However, we must ensure that we only add this character if it's different from the one we've selected from the start. This situation can arise when the string's length is odd, and we encounter the middle character. Therefore, we only append it if length - 1 - i is not equal to i.

Incorporating these steps, our function evolves to:

Kotlin
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1fun solution(inputString: String): String {
2    var result = ""
3    val length = inputString.length
4    for (i in 0 until length / 2 + length % 2) {
5        result += inputString[i]
6        if (length - 1 - i != i) {
7            result += inputString[length - 1 - i]
8        }
9    }
10    return result
11}
12
13fun main() {
14    // Example usage
15    val inputString = "abcdefg"
16    val outputString = solution(inputString)
17    println(outputString)  // Outputs: agbfced
18}

And there you have it, our function is now complete!

Congratulations! In this lesson, you have mastered a unique aspect of string manipulation. Although it may have appeared challenging, your perseverance has indeed paid off! Now, the process of selecting characters from a string in such an unusual pattern should feel intuitive.

Now it's time for some practice. As the adage goes, 'practice makes perfect,' and it holds true here as well. The most effective way to cement your understanding of a concept is to apply it. Challenge yourself with other problems that require similar techniques. Not only will this exercise embed these concepts in your mind, but it will also enhance your confidence in problem-solving. Enjoy coding, and remember, persistence is key!