Program to print window pattern

Difficulty Level : Easy
Last Updated : 22 Jan, 2021

Print the pattern in which there is a hollow square and plus sign inside it. The pattern will be as per the n i.e. number of rows given as shown in the example.

Examples:

Input : 6
Output : * * * * * *
         *   * *   * 
         * * * * * * 
         * * * * * * 
         *   * *   * 
         * * * * * *

Input : 7
Output : * * * * * * * 
         *     *     * 
         *     *     * 
         * * * * * * * 
         *     *     * 
         *     *     * 
         * * * * * * *

Approach:

We will start a for loop up till n and inside this also there is for loop up till n.
In this simply we have to check if the row is first or last or column is first or last, then print “*”.
Now we have to check for the middle row and column.
So when n is odd, we will have a middle row and column and if row or column is in middle then we will print “*”.
If n is even, then rows or columns if equal to these values n/2 and (n/2)+1, then we will print “*”.
Else everywhere we have to print ” “(space).

Below is the implementation.

C++14

// C++ program to print the pattern  
// hollow square with plus inside it 
// window pattern 
#include <bits/stdc++.h>
using namespace std; 
  
// Function to print pattern n means  
// number of rows which we want 
void window_pattern (int n)
{
    int c, d;
      
    // If n is odd then we will have 
    // only one middle element 
    if (n % 2 != 0) 
    {
        c = (n / 2) + 1;
        d = 0;
    }
  
    // If n is even then we will have two 
    // values 
    else
    {
        c = (n / 2) + 1;
        d = n / 2 ;
    }
  
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= n; j++)
        {
           
            // If i,j equals to corner row or 
            // column then "*" 
            if (i == 1 || j == 1 || 
                i == n || j == n)
                cout << "* ";
  
            else
            {
               
                // If i,j equals to the middle  
                // row or column then  "*" 
                if (i == c || j == c)
                    cout << "* ";
  
                else if (i == d || j == d)
                    cout << "* ";
  
                else
                    cout << "  ";
            }
        }
        cout << '\n';
    }
}
  
// Driver Code
int main()
{
    int n = 7;
   
    window_pattern(n);
    return 0;    
}
  
// This code is contributed by himanshu77

Java

// Java program to print the pattern  
// hollow square with plus inside it 
// window pattern 
class GFG
{
 
  // Function to print pattern n means  
  // number of rows which we want 
  static void window_pattern (int n)
  {
    int c, d;
 
    // If n is odd then we will have 
    // only one middle element 
    if (n % 2 != 0) 
    {
      c = (n / 2) + 1;
      d = 0;
    }
 
    // If n is even then we will have 
    // two values 
    else
    {
      c = (n / 2) + 1;
      d = n / 2 ;
    }  
    for(int i = 1; i <= n; i++)
    {
      for(int j = 1; j <= n; j++)
      {
 
        // If i,j equals to corner row 
        // or column then "*" 
        if (i == 1 || j == 1 || 
            i == n || j == n)
          System.out.print("* ");           
        else
        {
 
          // If i,j equals to the middle  
          // row or column then  "*" 
          if (i == c || j == c)
            System.out.print("* ");
          else if (i == d || j == d)
            System.out.print("* ");
          else
            System.out.print("  ");
        }
      }
      System.out.println();
    }
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int n = 7;
    window_pattern(n);
  }
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 program to print the pattern 
# hollow square with plus inside it
# window pattern
 
 
# function to print pattern n means 
# number of rows which we want
def window_pattern(n):
     
    # if n is odd then we will have
    # only one middle element
    if n % 2 != 0:
        c = ( n // 2 ) + 1
        d = 0
         
    # if n is even then we will have two
    # values
    else:
        c = ( n // 2 ) + 1
        d = ( n // 2 )
 
    for i in range( 1 , n + 1 ):
        for j in range( 1 , n + 1 ):
             
            # if i,j equals to corner row or
            # column then "*"
            if i == 1 or j == 1 or i == n or j == n:
                print("*",end=" ")
                 
            else:
                 
                # if i,j equals to the middle row 
                # or column then  "*"
                if i == c or j == c:
                    print("*",end=" ")
                     
                elif i == d or j == d:
                    print("*",end=" ")
                 
                else:
                    print(" ",end=" ")
         
        print()
 
 
# Driver Code 
if __name__ == "__main__":
    n = 7
    window_pattern(n)

C#

// C# program to print the pattern  
// hollow square with plus inside it 
// window pattern
using System;
 
class GFG{
 
// Function to print pattern n means  
// number of rows which we want 
static void window_pattern (int n)
{
    int c, d;
     
    // If n is odd then we will have 
    // only one middle element 
    if (n % 2 != 0) 
    {
        c = (n / 2) + 1;
        d = 0;
    }
   
    // If n is even then we will have 
    // two values 
    else
    {
        c = (n / 2) + 1;
        d = n / 2 ;
    }
   
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= n; j++)
        {
             
            // If i,j equals to corner row 
            // or column then "*" 
            if (i == 1 || j == 1 || 
                i == n || j == n)
                Console.Write("* ");
                 
            else
            {
                 
                // If i,j equals to the middle  
                // row or column then  "*" 
                if (i == c || j == c)
                    Console.Write("* ");
   
                else if (i == d || j == d)
                    Console.Write("* ");
   
                else
                    Console.Write("  ");
            }
        }
        Console.WriteLine();
    }
}
 
// Driver code
static void Main()
{
    int n = 7;
     
    window_pattern(n);
}
}
 
// This code is contributed by divyesh072019

Output :

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

Time complexity: O(n²)

Attention geek! Strengthen your foundations with the Python Programming Foundation Course and learn the basics.

To begin with, your interview preparations Enhance your Data Structures concepts with the Python DS Course.

My Personal Notes

Related Articles

C++14

Java

Python3

C#