Btukfyl

How is it possible to print the following pattern?

    *

   **

  ***

 ****

*****

****

***

**

*

How can we add the spaces during the first half of the pattern?
I only managed to get the second half of the pattern right:

#include <iostream>



using namespace std;



void printpattern(int n)

{

    for (int r = 0; r <= n; r++)

    {

        for (int z = 0; z <= r; z++) {

            cout << "*";

        }



        cout << endl;

    }

}



int main()

{

    int n = 5;

    printpattern(n);

}

Although comments suggest using std::string, I believe this was intended to be written using only loops. This works:

void printpattern(int n)

{

  // print first half.

  for (int i = 0; i < n; ++i) {

    // print spaces

    for (int r = n - i; r > 0; --r)

      std::cout << ' ';



    // print stars.

    for (int j = i; j > 0; --j) 

      std::cout << '*';



    std::cout << 'n';

  }



  /// print second half. No need to print spaces here.

  for (int i = 1; i <= n; ++i) {

    for (int r = n - i; r >= 0; --r) 

      std::cout << '*';



    std::cout << 'n';

  }

}

If one is afraid of using loops such thingies can of course always be solved with recursion and a little math:

#include <iostream>



void pattern(int n, int p = 0)

{   

    if (!n) return;

    if (!p) { pattern(2 * n * n - n, n); return; }

    int k = --n / p, o = n % p + 1, t = o - (p - k);

    std::cout.put(" *"[k >= p && t < p || k < p && t >= 0]);

    --o || std::cout.put('n');

    pattern(n, p);

}

It's easier to do this by composing a longer string and using a sliding view.

That's pretty straightforward in C++:

#include <iostream>

#include <string>

#include <string_view>



void printpattern(std::size_t n)

{

    const auto s = std::string(n, ' ') + std::string(n, '*') + std::string(n, ' ');

    for (std::size_t i = 1;  i < n*2;  ++i)

        std::cout << std::string_view(s.data()+i, n) << 'n';

}



int main()

{

    printpattern(5);

}

You could of course make the space padding be of length n-1 on both sides, and use a more conventional loop starting i at zero:

    const auto s = std::string(n - 1, ' ') + std::string(n, '*')

                 + std::string(n-1, ' ');

    for (std::size_t i = 0;  i < n * 2 - 1;  ++i)

It's up to you whether saving two characters of temporary string is worthwhile.