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)
Article Tags :
Recommended Articles