In the world of computer science and programming, searching for a specific element in a sorted array is a common task. This requires an efficient algorithm to quickly locate the desired element and its corresponding index. But what happens when the element we are searching for is equal to its index value? In this case, we need to find the index i where X[i] = i.

To better understand this scenario, let's first define what a sorted array is. A sorted array is an array in which the elements are arranged in a particular order, usually in ascending or descending order. This arrangement allows for faster search and retrieval of elements, making it a preferred data structure for many applications.

Now, let's dive into the problem at hand. Imagine we have an array of integers, X, in which each element is unique and sorted in ascending order. Our goal is to find the index i where X[i] = i. In simpler terms, we need to find the position of an element that is equal to its value.

One way to approach this problem is to use a linear search algorithm, which sequentially checks each element in the array until a match is found. However, this approach would have a time complexity of O(n), where n is the size of the array. This means that as the size of the array increases, the time taken to find the desired index also increases, making it an inefficient solution.

To improve the time complexity, we can use a binary search algorithm. This algorithm takes advantage of the fact that the array is sorted and employs a divide and conquer approach to search for the element. It starts by comparing the middle element of the array with the desired element and then divides the array into two halves based on the comparison. If the middle element is greater than the desired element, the search continues in the first half of the array. Similarly, if the middle element is less than the desired element, the search continues in the second half of the array. This process is repeated until the element is found or the entire array is exhausted.

Using this algorithm, we can find the desired index in O(log n) time complexity, which is a significant improvement over the linear search algorithm. This is because, with each comparison, the search space is divided into half, reducing the number of elements to be searched in the next iteration.

Let's take an example to understand this better. Consider the array X = [0, 2, 4, 6, 8, 10]. Here, the desired index i is 0, as X[0] = 0. Using the linear search algorithm, we would have to compare each element in the array until we find a match, resulting in a time complexity of O(n). On the other hand, with the binary search algorithm, we would start by comparing the middle element, 6, with the desired element, 0. Since 0 < 6, we continue the search in the first half of the array, [0, 2, 4]. Again, we compare the middle element, 2, with the desired element, 0. Since 0 < 2, we continue the search in the first half of the array, [0]. Finally, we find the desired index i = 0, with just three comparisons, making it a more efficient solution.

In conclusion, finding the index i where X[i] = i in a sorted array can be achieved using a binary search algorithm in O(log n) time complexity. This approach is significantly faster than the linear search algorithm and is widely used in various applications. So, the next time you come across a similar problem, remember to use a binary search algorithm for an efficient solution.