Question

Binary search is performed on a sorted array of n elements. The search key is not in the array and falls between the elements at positions m and m+1 (where 1 ≤ m < n). How many comparisons are needed in the worst case scenario to determine that the key is not in the array?

a. ceil(log(n))

b. floor(log(n))

c. ceil(log(m))

d. floor(log(m))

Answer 1

The search key is not in the array and falls between the elements at positions m and m+1 (where 1 ≤ m < n).
This statement describes a scenario where you are performing a binary search on a sorted array, and the search key is not present but its value lies between two consecutive elements in the array.

Answer 2

the answer is a ceil(o(logn))  because it’s just asking the worst case complexity of binary search and try to make students confuse , also ans is always in a ceil value not float remember