if i used reserve(n) would it fix that?

by initializing 'z' with no elements, the program is going to resize the vector during your loop multiple times which means allocating memory again and moving all the data over there (expensive). this happens every time you do a push_back and the capacity of your vector has been reached.

The kata description assures neither x nor n can be zero :)
You can declare i inside the for loop header.