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 = 9pattern(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)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
… [Trackback]
[…] Information on that Topic: geeksforgeeks.org/pattern-to-print-x-in-a-rectangular-box/ […]