Algorithm Interview Questions and How to be good at them.

I was asked to White Board Binary Search Algorithm, Read on if you want to get insight in how to be really well at cracking them.

Even though I would never test anyone on White Board and this interview practice is slowly diminishing, I believe there are few traditional companies who are still going this route.

A binary search algorithm finds an item in a sorted list in O(lg(n)) time.

A brute force search would walk through the whole list, taking O(n)O(n) time in the worst case.

Let’s say we have a sorted list of numbers. To find a number with a binary search, we:

We can do this recursively, or iteratively. Here’s an iterative version:

CC#C++JavaJavaScriptObjective-CPHPPython 2.7Python 3.6 (beta)RubySwift

So the question is, “how many times must we divide our original list size (nn) in half until we get down to 1?”

n * \frac{1}{2} * \frac{1}{2} * \frac{1}{2} * \frac{1}{2} * … = 1n∗21​∗21​∗21​∗21​∗…=1

How many \frac{1}{2}21​’s are there? We don’t know yet, but we can call that number xx:

n * (\frac{1}{2})^x = 1n∗(21​)x=1

Now we solve for xx:

n * \frac{1^x}{2^x} = 1n∗2x1x​=1n * \frac{1}{2^x} = 1n∗2x1​=1\frac{n}{2^x} = 12xn​=1n = 2^xn=2x

Now to get the xx out of the exponent. How do we do that? Logarithms.

The right hand side asks, “what power must we raise 22 to, to get 2^x2x?” Well, that’s just xx.

Careful: we can only use binary search if the input list is already sorted.

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