how to calculate binary search complexity

Question

I heard somebody say that since binary search halves the input required to search hence it is log(n) algorithm. Since I am not from a mathematics background I am not able to relate to it. Can somebody explain it in a little more detail? does it have to do something with the logarithmic series?

Answer 1

Here a more mathematical way of seeing it, though not really complicated. IMO much clearer as informal ones:

The question is, how many times can you divide N by 2 until you have 1? This is essentially saying, do a binary search (half the elements) until you found it. In a formula this would be this:

1 = N / 2^x

multiply by 2^x:

2^x = N

now do the log₂:

log₂(2^x)    = log₂ N
x * log₂(2) = log₂ N
x * 1         = log₂ N

this means you can divide log N times until you have everything divided. Which means you have to divide log N ("do the binary search step") until you found your element.