time complexity

Hi I have a question that:

consider I have T(n) = m * n^2 (n<m) is this correct to write T(n) = O(m) ? because I have written T(n) = m*n*n So bec开发者_开发技巧ause n<m we have T(n) = O(m) thanks

No, the best thing you can do is to write T(n,m) = O(m^3). n < m is a very weak condition and basically just gives you n in O(m). For example, n could always be m-1.

Edit: My first answer was imprecise, as T was only a function in n. If m is constant, the answer is still holds, but O(m^3) is equal to O(1).

I believe in this case T(n) = O(n^2)

The formal definition of big-O:

f(x) = O(g(x)) if and only if there exists a positive real number M and a real number x0 such that |f(x)| ≤ M|g(x)| for all x > x0.

In your case, T(n) will always be less than or equal to m(n^2).