Question

I understand how to find the time complexity of an algorithm when I’ve been presented with an algorithm, but I can’t seem to get my head around how to work it out when I’ve been given the number of times the algorithm is executed, and the time taken.

I can sometimes get it, when it’s obvious things like O(n), O(n) or O(n^2) but take this question for example:

An algorithm runs a given input of size n.
If n is 4096 the run time is 512 milliseconds.
If n is 16384 the run time is 1024 milliseconds.
If n is 36864 the run time is 1536 milliseconds.

What is the time complexity?

I see that as n * 2, t * 1.5, but I’m not quite sure how to work it out.

Thank you for your help 🙂

Answer 1

I would say you need more than just three datapoints for this kind of question, because of complexities in the system rather than just the algorithm.

What I would do is compare the iterations and the elapsed time and see if you can find a pattern that matches one of the standard time complexities:

• Constant: O(c) where c is a constant.
• Linear: O(n)
• Polynomial: O(n^c) where c is a constant (or even something complicated like O(n^2 + n^6))
• Exponential: O(c^n) where c is a constant.
• Logarithmic: O(log|n|)
• Whatever this is called: O(n log|n| )

Let’s go through your problem:

n     | time
4096  | 512 ms
16384 | 1024 ms
36864 | 1536 ms

When n goes up by a factor of 4 (from 4096 to 16384), the time goes up by a factor of 2 (from 512 to 1024 ms).

When n goes up by a factor of 9 (from 4096 to 36864), the time goes up by a factor of 3 (from 512 to 1536 ms).

A function that matches this is f(n) = n^(1/2). When n goes up by a factor of 4, f(n) goes up by a factor of sqrt(4), etc…

So this is of order O(n^.5), which is polynomial

TLDR: Graph it out and match it to a common function for time-complexity. In the real-world you would probably need more than three datapoints.

edit: I’d like to add that that this should really be more complex. There is likely a constant term in every kind of time complexity. I.e. O(n^c) is more likely O(n^c + K) where K is a constant. We ignore the constant for simplicity when writing this out, but it would show up in your graph.