The recurrence relation of binary search is (in the worst case)
T(n) = T(n/2) + O(1)
Using Master's theorem

- n is the size of the problem.
- a is the number of subproblems in the recursion.
- n/b is the size of each subproblem. (Here it is assumed that all subproblems are essentially the same size.)
- f (n) is the cost of the work done outside the recursive calls, which includes the cost of dividing the problem and the cost of merging the solutions to the subproblems.
Here a = 1, b = 2 and f(n) = O(1) [Constant]
We have f(n) = O(1) = O(nlogba)
=> T(n) = O(nlogba log2 n)) = O(log2 n)