If a list of elements is already sorted in ascending order, how many comparisons are needed in the worst-case scenario to find an element using Binary Search?

Binary Search is an efficient search algorithm that works on the principle of divide and conquer. It is used to search for an element in a sorted list.

Here's how it works:

1. It starts by comparing the target value to the middle element of the list.
2. If the target value matches the middle element, its position in the list is returned.
3. If the target value is less than the middle element, the search continues in the lower half of the list.
4. If the target value is greater than the middle element, the search continues in the upper half of the list.
5. This process continues until the target value is found or the sublist becomes empty.

In the worst-case scenario, the target value is not in the list. In this case, the number of comparisons needed is equal to the number of times we can halve the list until it becomes empty.

This is given by the formula log2(n) + 1, where n is the number of elements in the list. The "+1" is because in the worst case, we have to check the last remaining element to confirm that the target value is not in the list.

So, if a list of elements is already sorted in ascending order, the worst-case scenario for finding an element using Binary Search would require log2(n) + 1 comparisons.