print the pyramidical star pattern in C++

By Manisha Rathore

In this tutorial, we will learn to print the pyramidical star pattern in C++.

For Example:

• If n=5.
• The output:

* 
* * 
* * * 
* * * * 
* * * * *

Approach:

• Firstly, it is a star pattern.
• Now, we have to write the printing function to demonstrate this pattern.
• There will be two for loops.
• The outer for loop will handle the number of rows in that pattern.
• And, the inner loop will handle the number of columns in that pattern.
• The inner loop will depend upon outer for loop and we can say, it will change the values according to outer for loop.
• Finally, it will give you a pyramidical star pattern.

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print the pyramidical star pattern in C++

Hence, below is the iterative implementation of the above approach.

#include <iostream> 
using namespace std; 

 
void pyp(int n) 
{ 
  // outer loop  
  for (int i=0; i<n; i++) 
  { 
    // inner for loop will handle number of columns and their values changes acc. to outer for loop 
    for(int j=0; j<=i; j++ ) 
    { 
      // print star 
      cout << "* "; 
    } 

     
    cout << endl; 
  } 
} 


int main() 
{ 
  int n = 6; 
  pyp(n); 
  return 0; 
} 

OUTPUT EXPLANATION:

INPUT: n=6
OUTPUT:

* 
* * 
* * * 
* * * * 
* * * * * 
* * * * * *

After 180 degrees rotation: pyramidical star pattern in C++

Approach:

• Firstly, set k equal to the number of spaces that is approximate twice the value of n.
• Now, the change is, there is three for loops.
• Then, the outer loop will handle the number of rows.
• The first inner loop will handle the number of spaces that will change according to the outer for loop.
• Here, decrement value of k by 2 after each first inner loop.
• Now, the second for loop will handle the number of columns that will change according to the outer for loop.
• Finally, print the stars according to the above approach.

Hence implementation is here:

 
#include <iostream> 
using namespace std; 
 
void pyp2(int nu) 
{  
  int k = 2*nu - 2; 

  
  for (int i=0; i<nu; i++) 
  { 
 
    for (int j=0; j<k; j++) 
      cout <<" "; 
    k = k - 2; 

     for (int j=0; j<=i; j++ ) 
    { 
      // prints 
      cout << "* "; 
    } 

    
    cout << endl; 
  } 
} 

 
int main() 
{ 
  int nu = 6; 
  pyp2(nu); 
  return 0; 
}

INPUT: nu=6
OUTPUT:

          * 
        * * 
      * * * 
    * * * * 
  * * * * * 
* * * * * *