Hello and welcome to our exciting exploration of C++ 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 C++ function, std::string solution(std::string inputString). This function takes inputString as a parameter, a string of lowercase English alphabet letters ('a' to 'z'), with a length ranging between 1 to 100characters. 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.

C++
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1std::string solution(std::string inputString) {
2    std::string result = "";

Upon initialization, we need to traverse the inputString. C++ furnishes a for loop to iterate over all elements in an array 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 inputString.length() / 2 + 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:

C++
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1std::string solution(std::string inputString) {
2    std::string result = "";
3    int length = inputString.length();
4    for (int i = 0; i < length / 2 + length % 2; i++) {
5        // Implementation in next step
6    }

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 make certain 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:

C++
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1std::string solution(std::string inputString) {
2    std::string result = "";
3    int length = inputString.length();
4    for (int i = 0; i < length / 2 + length % 2; i++) {
5        result += inputString[i];
6        if (length - 1 - i != i) {
7            result += inputString[length - 1 - i];
8        }
9    }
10    return result;
11}

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

Congratulations! In this lesson, you have mastered a unique aspect of string manipulation in C++. 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 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, remember, persistence is key!