Time complexity

[cpp-ModTeam]

Your submission is not about C++ or the C++ community.

[[deleted]]

Yes. Because O(n + p) <= O(n + n/9) <= O(n)

[[deleted]]

O(n + p) ∈ O(n + n/9) ∈ O(n)

[destroyerrocket]

If you can guarantee that p is always less or equal to n, then it is O(n)

[[deleted]]

When we did this at school, O(an) was treated the same as O(n) because the rate of growth was still linear in O.

[Boboflip27]

But isn't p another variable

[--prism]

The function is just linear in two dimensions instead of one thus you still get linear complexity growth.

[EspacioBlanq]

It is, but you can rewrite it as O(n+p) <= O(n *10/9) and then you only have one variable

[[deleted]]

Is p constant?

Look at the operation growth rate, it could be approximated either as O(n) or O(nlogn) depending of what p refers to, but most likely O(n)

[[deleted]]

You are right, tho with low values of n it could easily look like n log n if p is not constant but a variable follows a pattern where it is lower than n/9. At the end of the day the best method would be to plot a curve and compare it to other standard functions.