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Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and a supportive community. Start your free trial today. # Polynomial Runtime (O(n^k)) vs Exponential Runtime(O(k^n))

How do you choose n and k when determining if an algorithm's complexity is of O(n^k) or O(k^n)? In the padlock example, when the size is 2, there're 10^2 possible combinations. Isn't this running in quadratic time?

10^2 seems "quadratic-y", but notice that the varying variable is not the base (10), but is the exponent!

n^2 (a quadratic) looks like this for the varying `n`s:

```1^2   2^2   3^2   4^2   5^2   ...
```

But the exponential 10^n look like:

```10^1   10^2   10^3   10^4   10^5   ...
```

Notice how the exponent, not the base, is varying.

Exponentials explode off even more than quadratics/cubics, so generally it is very bad to have an exponential algorithm. Thank you for clarifying how an exponential algorithm can grow more than a polynomial one, but I still have the question: How do you know if n or k should be the varying variable? How do you choose which is which?

Usually `n` is the varying variable, and `k` refers to some constant Hello James Estrada, I hope this helps:

There are 2 boxes to be filled using 10 numbers(0-9).
number of combinations =(number of ways in which the box1 can be filled)x(number of ways in which box 2 can be filled)...(number of ways in which the box n can be filled)
either box can be filled by 0 or 1 or 2 ....or 9 (10 ways).
therefore, total number of combinations is (10) X (10) or 10^2.
when number of boxes increase to 3, '2' changes to '3' but the range of numbers will still be (0-9). So, '10' does not change. (the answer would then be 10 x 10 x 10 = 10^3)
So, k=10(the number of items in range ) and n=2(number of boxes)