Solid square inside a hollow square | Pattern
Last Updated :
20 Feb, 2023
Given the value of n, print a hollow square of side length n and inside it a solid square of side length (n – 4) using stars(*).
Examples :
Input : n = 6
Output :
******
* *
* ** *
* ** *
* *
******
Input : n = 11
Output :
***********
* *
* ******* *
* ******* *
* ******* *
* ******* *
* ******* *
* ******* *
* ******* *
* *
***********
C++
#include <bits/stdc++.h>
using namespace std;
void print_Pattern( int n)
{
for ( int i = 1; i <= n; i++) {
for ( int j = 1; j <= n; j++) {
if ((i == 1 || i == n) || (j == 1 || j == n) ||
(i >= 3 && i <= n - 2) && (j >= 3 && j <= n - 2))
cout<< "*" ;
else
cout<< " " ;
}
cout<<endl;
}
}
int main()
{
int n = 6;
print_Pattern(n);
return 0;
}
|
Java
import java.io.*;
class GFG
{
static void print_Pattern( int n)
{
for ( int i = 1 ; i <= n; i++)
{
for ( int j = 1 ; j <= n; j++)
{
if ((i == 1 || i == n) || (j == 1 || j == n) ||
(i >= 3 && i <= n - 2 ) && (j >= 3 && j <= n - 2 ))
System.out.print( "*" );
else
System.out.print( " " );
}
System.out.println();
}
}
public static void main (String[] args)
{
int n = 6 ;
print_Pattern(n);
}
}
|
Python3
def print_Pattern(n):
for i in range ( 1 , n + 1 ):
for j in range ( 1 , n + 1 ):
if ((i = = 1 or i = = n) or
(j = = 1 or j = = n) or
(i > = 3 and i < = n - 2 ) and
(j > = 3 and j < = n - 2 )):
print ( "*" , end = "")
else :
print (end = " " )
print ()
n = 6
print_Pattern(n)
|
C#
using System;
class GFG {
static void print_Pattern( int n)
{
for ( int i = 1; i <= n; i++)
{
for ( int j = 1; j <= n; j++)
{
if ((i == 1 || i == n) ||
(j == 1 || j == n) ||
(i >= 3 && i <= n - 2) &&
(j >= 3 && j <= n - 2))
Console.Write( "*" );
else
Console.Write( " " );
}
Console.WriteLine();
}
}
public static void Main ()
{
int n = 6;
print_Pattern(n);
}
}
|
PHP
<?php
function print_Pattern( $n )
{
for ( $i = 1; $i <= $n ; $i ++)
{
for ( $j = 1; $j <= $n ; $j ++)
{
if (( $i == 1 || $i == $n ) ||
( $j == 1 || $j == $n ) ||
( $i >= 3 && $i <= $n - 2) &&
( $j >= 3 && $j <= $n - 2))
echo "*" ;
else
echo " " ;
}
echo "\n" ;
}
}
$n = 6;
print_Pattern( $n );
?>
|
Javascript
<script>
function print_Pattern(n)
{
for ( var i = 1; i <= n; i++)
{
for ( var j = 1; j <= n; j++)
{
if (
i == 1 ||
i == n ||
j == 1 ||
j == n ||
(i >= 3 && i <= n - 2 && j >= 3 && j <= n - 2)
)
document.write( "*" );
else document.write( " " );
}
document.write( "<br>" );
}
}
var n = 6;
print_Pattern(n);
</script>
|
Output :
******
* *
* ** *
* ** *
* *
******
Time complexity: O(n2) for given input n
Auxiliary Space: O(1)
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