Hello and welcome to our exciting exploration of JavaScript 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. However, 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, then 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 JavaScript function:

function solution(inputString) {
    // function implementation
}

This function takes inputString as a parameter, a string of lowercase English alphabet letters ('a' to 'z'), with a length ranging between 1 and 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 array to stockpile the output.

function solution(inputString) {
    let result = [];

Upon initialization, we need to traverse the inputString. JavaScript 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 can realize this by employing Math.ceil(inputString.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:

function solution(inputString) {
    let result = [];
    let length = inputString.length;
    for (let i = 0; i < Math.ceil(length / 2); i++) {
        // Implementation in next step