Given the value of n, i.e, number of lines, print the pattern.
Examples :
Input: n = 5
Output:
*_*****_*****_*
**_****_****_**
***_***_***_***
****_**_**_****
*****_*_*_*****
Input: n = 10
Output:
*_**********_**********_*
**_*********_*********_**
***_********_********_***
****_*******_*******_****
*****_******_******_*****
******_*****_*****_******
*******_****_****_*******
********_***_***_********
*********_**_**_*********
**********_*_*_**********
C++
#include <bits/stdc++.h>
using namespace std;
void print1( int i)
{
for ( int j = 1; j <= i; j++)
cout << "*" ;
}
void print( int n)
{
for ( int i = 1; i <= n; i++) {
print1(i);
cout << "_" ;
print1(n - i + 1);
cout << "_" ;
print1(n - i + 1);
cout << "_" ;
print1(i);
cout << endl;
}
}
int main()
{
int n = 5;
print(n);
return 0;
}
|
Java
import java.io.*;
class GFG
{
static void print1( int i)
{
for ( int j = 1 ; j <= i; j++)
System.out.print( "*" );
}
static void print( int n)
{
for ( int i = 1 ; i <= n; i++) {
print1(i);
System.out.print( "_" );
print1(n - i + 1 );
System.out.print( "_" );
print1(n - i + 1 );
System.out.print( "_" );
print1(i);
System.out.println();
}
}
public static void main (String[] args)
{
int n = 5 ;
print(n);
}
}
|
Python3
def print1(i):
for j in range ( 1 , i + 1 ):
print ( "*" , end = "")
def printPattern(n):
for i in range ( 1 , n + 1 ):
print1(i)
print ( "_" , end = "")
print1(n - i + 1 )
print ( "_" , end = "")
print1(n - i + 1 )
print ( "_" , end = "")
print1(i)
print ("")
n = 5
printPattern(n)
|
C#
using System;
class GFG {
static void print1( int i)
{
for ( int j = 1; j <= i; j++)
Console.Write( "*" );
}
static void print( int n)
{
for ( int i = 1; i <= n; i++) {
print1(i);
Console.Write( "_" );
print1(n - i + 1);
Console.Write( "_" );
print1(n - i + 1);
Console.Write( "_" );
print1(i);
Console.WriteLine();
}
}
public static void Main ()
{
int n = 5;
print(n);
}
}
|
PHP
<?php
function print1( $i )
{
for ( $j = 1; $j <= $i ; $j ++)
echo "*" ;
}
function printp( $n )
{
for ( $i = 1; $i <= $n ; $i ++)
{
print1( $i );
echo "_" ;
print1( $n - $i + 1);
echo "_" ;
print1( $n - $i + 1);
echo "_" ;
print1( $i );
echo "\n" ;
}
}
$n = 5;
printp( $n );
?>
|
Javascript
<script>
function print1(i)
{
for ( var j = 1; j <= i; j++)
document.write( "*" );
}
function print(n)
{
for ( var i = 1; i <= n; i++) {
print1(i);
document.write( "_" );
print1(n - i + 1);
document.write( "_" );
print1(n - i + 1);
document.write( "_" );
print1(i);
document.write( '<br>' );
}
}
var n = 5;
print(n);
</script>
|
Time complexity: O(n2), where n is the given integer.
Auxiliary space: O(1) using constant space for variables