Welcome to today's lesson! We're diving into Binary Search, a clever technique for locating specific elements within a sorted list. We can find the targeted item by repeatedly dividing the search interval in half. It's akin to flipping through a dictionary — instead of going page by page, you'd start in the middle, then narrow down the section in half until you find your desired word.

Binary Search begins at the midpoint of a sorted list, halving the search area at each step until it locates the target. For example, if you're looking for the number 8 in a sorted list ranging from 1 to 10, you would start at 5. Since 8 is larger than the midpoint, you narrow the search to the second half of the list, leaving you with numbers 6 to 10. In this new sublist, the middle number is 8, and thus, you've found your target. This efficient approach significantly reduces the number of comparisons needed compared to a linear search.

Let's see how Binary Search can be implemented in Scala, taking a recursive approach. This process involves a function calling itself — with a base case in place to prevent infinite loops — and a recursive case to solve smaller parts of the problem.

def binarySearch(arr: Array[Int], start: Int, end: Int, target: Int): Int = {
    if (start > end) return -1 // Base case

    val mid = start + (end - start) / 2 // Find the midpoint

    if (arr(mid) == target) return mid // Target found

    if (arr(mid) > target) // If the target is less than the midpoint
        binarySearch(arr, start, mid - 1, target) // Search the left half
    else
        binarySearch(arr, mid + 1, end, target) // Search the right half
}

In this Scala code, the base case is defined first. If the start index is greater than the end index, it indicates the search area is exhausted, resulting in a -1 return. The code then locates the midpoint. If the midpoint equals our target, it’s returned. Depending on whether the target is less than or greater than the midpoint, the search continues within the left or right half, respectively.

Let's analyze the time complexity of Binary Search, which measures how much time an algorithm takes as the input size increases. Notably, Binary Search halves the list at every step, necessitating log(n) steps for an array of size n. Therefore, the time complexity of Binary Search is O(log n).