Binary search is also an efficient mechanism for only a sorted item list. It separates the element into one half of the checklist repeatedly, till you have narrowed it down to the only single place.
In this search, the number of comparisons is much more efficient than in the series to use the binary search approach, the items also should be shown, that's why the last choice is correct.
Binary search will start in the middle of the list, find which side the value is at, then look at the middle of those values. This is incredibly efficient for larger values (Although it needs to be sorted) and practical in many instances. Let's say that the desired number is 1, the binary search would look at "50" first, then proceed to compare with "5" and finally on it's third iteration, pick 1.
