This is a subject many people are afraid of, or simply don't get. At first, it seems to be mystical hocus-pocus, but today I'll show you a simple way to quickly get an estimate of the Big O of an algorithm/function. This is not the only way to determine Big O, nor do I make any claims of it being 100 percent accurate or effective. The idea here is to be able to look at loops, functions, and code in general, in a different light.

First, a definition:



The wiki article from which this is taken is an excellent reference if you need to quickly know the Big O of common/popular algorithms. What it absolutely fails to do is explain how one determines the actual Big O [I will use this term interchangeably with upper bound or limit]. For those not mathematically inclined, their eyes probably glaze over when reading the symbol infested portions. At running the risk of repeating myself, this article is simply here to show another way, one I think is easier (i.e. you could have a solid grasp on the concept without ever taking Calculus).

Some Basic Rules:

1. Nested loops are multiplied together.
2. Sequential loops are added.
3. Only the largest term is kept, all others are dropped.
4. Constants are dropped.
5. Conditional checks are constant (i.e. 1).

That's it really. I used the word loop, but the concept applies to conditional checks, full algorithms, etc.. since a whole is the sum of its parts. I can see the worried look on your face, this would all be frivolous without some examples[see code comments]:



Here we iterate 'n' times. Since nothing else is going on inside the loop (other then constant time printing), this algorithm is said to be O(n). The common bubble-sort:



Each loop is 'n'. Since the inner loop is nested, it is n*n, thus it is O(n^2). Hardly efficient. We can make it a bit better by doing the following:



Outer loop is still 'n'. The inner loop now executes 'i' times, the end being (n-1). We now have (n(n-1)/2). This is still in the bound of O(n^2), but only in the worst case.

An example of constant dropping:


At first you might say that the upper bound is O(2n); however, we drop constants so it becomes O(n). Mathematically, they are the same since (either way) it will require 'n' elements iterated (even though we'd iterate 2n times).

An example of sequential loops:



You wouldn't do this exact example in implementation, but doing something similar certainly is in the realm of possibilities. In this case we add each loop's Big O, in this case n+n^2. O(n^2+n) is not an acceptable answer since we must drop the lowest term. The upper bound is O(n^2). Why? Because it has the largest growth rate (upper bound or limit for the Calculus inclined).

Finite loops are common as well, an example:



Outer loop is 'n', inner loop is 2, this we have 2n, dropped constant gives up O(n).

In short Big O is simply a way to measure the efficiency of an algorithm. The goal is constant or linear time, thus the various data structures and their implementations. Keep in mind that a "faster" structure or algorithm is not necessary better. For example, see the classic hash table versus binary tree debate. While not 100% factual, it often said that a a hash-table is O(1) and is therefore better then a tree. From a discussion on the subject in a recent class I took:



The above is always something good to keep in mind when dealing with theoretical computer science concepts. Hopefully you found this both interesting and helpful. Happy coding!

I noticed I didn't provide a log(n) or nlog(n) example, arguably the toughest ones.

A quick metric to see if a loop is log n is to see how the counter increments in relationship to the total number of elements.

Example:




There are n iterations, however, instead of simply incrementing, 'i' is increased by 2*itself each run. Thus the loop is log(n).

An example of nested loops:



This example is n*log(n). (Remember that nested loops multiply their Big O's.)

Great tutorial Kya! Thanks a million!

This post has been edited by Dogstopper: 20 Sep, 2009 - 08:49 AM

Very helpful.

Thanks KYA.

A Big O thanks KYA for your tutotial. I am a novice in this field and have got lot from this post. will you please simplify Big O notation for the function 'logn + 3n'. Got confused from two different books one is mentioning it as O(logn) and the other O(n).

And one more thing is that don't you think the above code will run for infinite times as the value of i will never incremented by i *=2, because everytime i=0. Will it be like :



If I am wrong, please explain.

The top one will run infinitely.


The bottom one is logn since i is incremented by twice itself each iteration.


O(n) would be if 'i' is incremented by one each time.