Print "*" up to n terms in pattern and its reverse - c++

Question

I have a problem that is asking for me to write a C++ program using for loops with less than 3 “cout” statements in your code to print the following pattern (ignore the pipes, the asterisks wouldn't appear without them):
|*
|***
|*****
|*******
|*********
|*********
|*******
|*****
|***
|*
This is my code I used for a fibonacci generator and I feel like it might be similar. I am able to print the "*" symbol but not in horizontal lines. What I need most help with is reversing the output. As in if given number n, I want the series to go n numbers into the series and then back down to 0.
#include <iostream>
using namespace std;
int main()
{
int y = 1, sum = 1, n;
cout << "Enter the number of terms you want" << endl;
cin >> n;
cout << "First " << n << " terms are :- " << endl;
for (int x = 0; x < n; x++) {
cout << "\n" <<endl;
for (int i = 0; i < sum; i++) {
cout << "*" << endl;
}
sum = y + 2;
y = sum;
}
}

Answer 1

It seems this is a homework, so I give some hints instead of a full solution.
For printing the *s in one line, please note that << endl will end the line in the output, i.e. print a line break. (The same does << "\n" by the way.) Not every cout statement has to have an << endl at its end.
For reversing the fibonacci sequence, once you have the last number in the variable sum, just do the reverse computation (i.e. subtraction). This could be done in a second set of loops, however, since you should not use cout statements too often, you better reuse the same loop by using some additional variable holding the current state (i.e. if you are counting up or down) and using an if to decide which computation to do. (I read the requirements such that only the cout statements for printing the pattern count to the "less than three" = 2)