Can a Big-Oh time complexity contain more than one variable?

Question

Let us say for instance I am doing string processing that requires some analysis of two strings. I have no given information about what their lengths might end up being, so they come from two distinct families. Would it be acceptable to call the complexity of an algorithm $O(n * m)$ or $O(n + m)$ (depending on if we use a naive or an optimized algorithm)?

On a similar vein, let us presume the algorithm we choose actually requires two stages - a setup phase on the first string which allows us to process any number of other strings without incurring that initial cost. Would it be considered appropriate to say it has a $O(n)$ construction followed by any number of $O(m)$ calculations?

Would it be appropriate to just call them $O(n)$ because both calculations are linear?

Answer 1

Yes, of course. This is fine and perfectly acceptable. It is common and standard to see algorithms whose running time depends upon two parameters.

For instance, you will often see the running time of depth-first search expressed as $O(n+m)$, where $n$ is the number of vertices and $m$ is the number of edges in the graph. This is perfectly valid. The meaning of this is there exists a constant $c$ and numbers $n_0,m_0$ such that the running time of the algorithm is at most $c \cdot (n+m)$, for all $n>n_0,m>m_0$. In other words, if the exact running time is $f(n,m)$, we say that $f(n,m) = O(n+m)$ if there exists $c,n_0,m_0$ such that $n>n_0$ and $m>m_0$ implies $f(n,m) \le c \cdot (n+m)$.

Yes, it is perfectly appropriate and acceptable to say that the first stage takes $O(n)$ time and the second stage takes $O(m)$ time.

Important: make sure you define what $n$ and $m$ are. You can't say "this is an $O(n)$ time algorithm" without specifying what $n$ is. If $n$ isn't specified in the problem statement, you need to specify it. For instance, see graph algorithms, where we typically define $n = $ # of vertices and $m = $ # of edges.

As far as whether you can call them $O(n)$ time, no, of course not -- unless you somehow know that $m = O(n)$. Of course, if you know that $m = O(n)$, then it follows that $m+n = O(n)$, so an $O(m+n)$ time algorithm is also an $O(n)$ time algorithm. But if there is no guarantee that $m = O(n)$, then you cannot call it an $O(n)$ time algorithm.

This is basic stuff. You'll find it all over algorithms textbooks.