# Why binary search offer O(log2N)?

Each comparision eliminates about half of the remaning items from consideration. What is the maximum number of comparisions this algorithm will require to check the entire list?

If we start with N items, about **N/2** items will be left after the first comparision.

After the 2nd comparision, there will be about **N/4**. Then **N/8**, **N/16** and so on.

When we split the list enough times, we end up with a list that has just 1 item. Either that is the item we're looking for or it is not. Either way, we're done.

The number of comparisions necessary to get to this point is **i** where **N/2^i = 1**. Solving for **i** give us **i = logN**. The maximum number of comparisions is logarithmic with respect to the number of items in the list.

Therefore the binary search is **O(logN)**.