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Pattern to print X in a rectangular box
  • Difficulty Level : Basic
  • Last Updated : 21 May, 2021

Given the value of length, print the X pattern in a box using # and ” “
Examples: 
 

Input  :   10 
Output :   ##########
           ##      ##
           # #    # #
           #  #  #  #
           #   ##   #
           #   ##   #
           #  #  #  #
           # #    # #
           ##      ##
           ##########

 

Input  :   7 
Output :   #######
           ##   ##
           # # # #  
           #  #  #
           # # # #
           ##   ##
           #######

Below is the implementation to print X in a rectangular box pattern :
 

C++




// CPP code to print the above
// specified pattern
#include <bits/stdc++.h>
using namespace std;
 
// Function to print pattern
void pattern(int n)
{
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (i == 0 || i == n - 1 ||
                j == 0 || j == n - 1 ||
                i == j || i == n - 1 - j)            
                cout << "#";           
 
            else
                cout << " ";
             
        }
 
        cout << endl;
    }
}
 
// Driver program
int main()
{
    int n = 9;
    pattern(n);
    return 0;
}

Java




// Java code to print the above
// specified pattern
class GFG {
     
    // Function to print pattern
    static void pattern(int n)
    {
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < n; j++)
            {
                if (i == 0 || i == n - 1 ||
                    j == 0 || j == n - 1 ||
                    i == j || i == n - 1 - j)            
                    System.out.print("#");        
     
                else
                    System.out.print(" ");
                 
            }
     
            System.out.println();
        }
    }
     
    // Driver Code
    public static void main(String args[])
    {
        int n = 9;
        pattern(n);;
         
    }
}
 
// This code is contributed by Sam007

Python 3




# Python3 code to print the above
# specified pattern
 
# Function to print pattern
def pattern(n):
 
    for i in range(0, n):
        for j in range(0, n):
            if (i == 0 or i == n - 1
                or j == 0 or j == n - 1
                or i == j or i == n - 1 - j):        
                print( "#", end="")
 
            else:
                print( " ",end="")
        print("")
         
# Driver program
n = 9
pattern(n)
 
# This code is contributed by Smitha.

C#




// C# code to print the above
// specified pattern
using System;
 
public class GFG {
 
    // Function to print pattern
    static void pattern(int n)
    {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (i == 0 || i == n - 1 ||
                    j == 0 || j == n - 1 ||
                    i == j || i == n - 1 - j)            
                    Console.Write("#");        
     
                else
                    Console.Write(" ");
                 
            }
     
            Console.WriteLine();
        }
    }
     
    // Driver code
    public static void Main()
    {
        int n = 9;
        pattern(n);
    }
}
 
// This code is contributed by Sam007.

PHP




<?php
// PHP code to print the above
// specified pattern
 
// Function to print pattern
function pattern($n)
{
    for ($i = 0; $i < $n; $i++) {
        for ($j = 0; $j < $n; $j++) {
            if ($i == 0 || $i == $n - 1 ||
                $j == 0 || $j == $n - 1 ||
                $i == $j || $i == $n - 1 - $j)            
                echo "#";        
 
            else
                echo " ";
        }
 
        echo "\n";
    }
}
 
    // Driver Code
    $n = 9;
    pattern($n);
     
// This code is contributed by nitin mittal
?>

Javascript




<script>
      // JavaScript code to print the above
      // specified pattern
      // Function to print pattern
      function pattern(n) {
        for (var i = 0; i < n; i++) {
          for (var j = 0; j < n; j++) {
            if (
              i === 0 ||
              i === n - 1 ||
              j === 0 ||
              j === n - 1 ||
              i === j ||
              i === n - 1 - j
            )
              document.write("#");
            else document.write("  ");
          }
 
          document.write("<br>");
        }
      }
 
      // Driver code
      var n = 9;
      pattern(n);
    </script>
Output: 
#########
##     ##
# #   # #
#  # #  #
#   #   #
#  # #  #
# #   # #
##     ##
#########

 




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