Exciting Exploration of C# Strings

Hello and welcome to our exciting exploration of C# strings! For today's lesson, we've prepared something quite unique: you will learn how to select characters from a string in an order that might initially seem a bit strange but is certainly interesting. We will explain it in such a comprehensive and concise way that you'll master it in no time. Let's get started!

Imagine this: You are given a string, and you need to go through it and pick its characters, but the order in which you pick them is unusual. You start with the first character, then jump to the last character, move to the second character, then to the second-to-last character, and carry on in this pattern until you run out of characters. Sounds interesting, doesn't it?

Here's what we mean:

You are tasked with writing a C# method, Solution(string inputString). This method will take inputString as a parameter, a string composed of lowercase English alphabet letters ('a' to 'z') with a length anywhere between 1 to 100 characters. The method will then return a new string, which is derived from the input string but with characters selected in the order we just discussed.

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

Before we begin problem-solving, let's set up our result store. We initialize a variable, result, as an empty string to store the output.

public class SolutionClass
{
    public string Solution(string inputString)
    {
        string result = "";
    }
}

After initialization, we need to iterate over the inputString. C# provides the for loop that we can use to define the number of times the loop should run.

The question here is, how many iterations are required for our loop? Given that we are taking two characters per iteration — one from the beginning and one from the end — we need the loop to run for half the length of the list if the length is even and half the length plus one if the length is odd so that we include the middle character.

How do we achieve this? We use length / 2 + length % 2. This will ensure that the loop iterates for half the length if it is even and half the length plus one if it is odd.

Here's what our method looks like so far:

public class SolutionClass
{
    public string Solution(string inputString)
    {
        string result = "";
        int length = inputString.Length;
        for (int i = 0; i < length / 2 + length % 2; i++)
        {
            // Loop implementation in next step
        }
        return result;
    }
}