How do you calculate the big oh of the binary search algorithm?

Question

I\'m looking for the mathematical proof, not just the answer.

Answer 1

The proof is quite simple: With each recursion you halve the number of remaining items if you’ve not already found the item you were looking for. And as you can only divide a number n recursively into halves at most log₂(n) times, this is also the boundary for the recursion:

2·2·…·2·2 = 2^x ≤ n ⇒ log₂(2^x) = x ≤ log₂(n)

Here x is also the number of recursions. And with a local cost of O(1) it’s O(log n) in total.