Time complexity of binary search
In binary search each step of the algorithm divides the list of items being searched in half of the list.
So we can say that
T(n)= T(n/2) +1 where n>1
= T(n/4) +1 +1
= T(n/22) + 2
= T(n/23) +3
……………..
………………
=T(n/2k) + k
If n=2k then k= log2n
= T(1) + log2n
So
T(n)= O(log2n)
Best case condition occur when the element to be searched in first time. Then complexity is O(1).