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.
The cost of `clear()` depends greately on what the stored objects are, and in particular whether they have a trivial destructor. If the type does not have a trivial destructor, then the call must destroy all stored objects and it is in fact an O(n) operation, but you cannot really do anything better.

Now, if the stored elements have trivial destructors, then the implementation can optimize the cost away and `clear()` becomes a cheap O(1) operation (just resetting the size --`end` pointer).

Remember that to understand asymptotic complexity you need to know what it talks about. In the case of `clear()` it represents the number of destructors called, but if the cost (hidden) is 0, then the operation is a no-op.