Question: I have a code where I routinely fill a vector with between 0 and 5000 elements. I know the maximum never exceeds 5000. Instead of initializing vector multiple times, I would like to do just once

```
vector<struct> myvector;
myvector.reserve(5000);

```

However, to fill the vector again, I have to clear the vector first without altering its capacity. So usually I call myvector.clear();

This is a O(n) operation. Is there something simple I can do to increase the performance of this or is this about the best that it will get?

Answer: 
If your struct has a non-trivial destructor, then that needs to be called for all the elements of the vector regardless of how it is emptied. If your struct only has a trivial destructor, the compiler or the standard library implementation is allowed to optimize away the destruction process and give you a O(1) operation.
Anything you do to remove the existing items from the vector needs to (potentially) invoke the destructor of each item being destroyed. Therefore, from the container's viewpoint, the best you can hope for is linear complexity.

That leaves only the question of what sort of items you store in the vector. If you store something like `int` that the compiler can/will know ahead of time has no destructor to invoke, chances are at least pretty good that removal will end up with constant complexity.

I doubt, however, that changing the syntax (e.g., `clear()` vs. `resize()` vs. `erase(begin(), end())` ) will make any significant difference at all. The syntax doesn't change that fact that (in the absence of threading) invoking N destructors is an O(N) operation.