Pattern to print X in a rectangular box

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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 
import java.io.*; 
  
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)              
                    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:

#########
##     ##
# #   # #
#  # #  #
#   #   #
#  # #  #
# #   # #
##     ##
#########

Time complexity: O(n^2) for given n
Auxiliary space: O(1)

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