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###### James Estrada

**Full Stack JavaScript**Techdegree Student 25,862 Points

# 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?

## 2 Answers

###### Alexander Davison

65,456 PointsNo is the short answer.

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.

###### G H Mahimaanvita

16,234 PointsHello 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)**

## James Estrada

Full Stack JavaScriptTechdegree Student 25,862 Points## James Estrada

Full Stack JavaScriptTechdegree Student 25,862 PointsThank 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?

## Alexander Davison

65,456 Points## Alexander Davison

65,456 PointsUsually

`n`

is the varying variable, and`k`

refers to some constant