Program to print the following pattern:
Examples :
Input : 5 Output: * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
This program is divided into four parts.
// C++ program to print // the given pattern #include<iostream> using namespace std;
void pattern( int n)
{ int i, j;
// This is upper half of pattern
for (i = 1; i <= n; i++)
{
for (j = 1; j <= (2 * n); j++)
{
// Left part of pattern
if (i > (n - j + 1))
cout << " " ;
else
cout << "*" ;
// Right part of pattern
if ((i + n) > j)
cout << " " ;
else
cout << "*" ;
}
cout << endl ;
}
// This is lower half of pattern
for (i = 1; i <= n; i++)
{
for (j = 1; j <= (2 * n); j++)
{
// Right Part of pattern
if (i < j)
cout << " " ;
else
cout << "*" ;
// Left Part of pattern
if (i <= ((2 * n) - j))
cout << " " ;
else
cout << "*" ;
}
cout << endl;
}
} // Driver Code int main()
{ pattern(7);
return 0;
} // This code is contributed by bunnyram19 |
// C program to print // the given pattern #include<stdio.h> void pattern( int n)
{ int i,j;
// This is upper half of pattern
for (i=1; i<=n; i++)
{
for (j=1; j<=(2*n); j++)
{
// Left part of pattern
if (i>(n-j+1))
printf ( " " );
else
printf ( "*" );
// Right part of pattern
if ((i+n)>j)
printf ( " " );
else
printf ( "*" );
}
printf ( "\n" );
}
// This is lower half of pattern
for (i=1; i<=n; i++)
{
for (j=1; j<=(2*n); j++)
{
// Right Part of pattern
if (i<j)
printf ( " " );
else
printf ( "*" );
// Left Part of pattern
if (i<=((2*n)-j))
printf ( " " );
else
printf ( "*" );
}
printf ( "\n" );
}
} // Driver Code int main()
{ pattern(7);
return 0;
} |
// Java program to print // the given pattern import java.io.*;
class GFG {
static void pattern( int n)
{
int i, j;
// This is upper half of pattern
for (i = 1 ; i <= n; i++) {
for (j = 1 ; j <= ( 2 * n); j++) {
// Left part of pattern
if (i > (n - j + 1 ))
System.out.print( " " );
else
System.out.print( "*" );
// Right part of pattern
if ((i + n) > j)
System.out.print( " " );
else
System.out.print( "*" );
}
System.out.println( "" );
}
// This is lower half of pattern
for (i = 1 ; i <= n; i++) {
for (j = 1 ; j <= ( 2 * n); j++) {
// Right Part of pattern
if (i < j)
System.out.print( " " );
else
System.out.print( "*" );
// Left Part of pattern
if (i <= (( 2 * n) - j))
System.out.print( " " );
else
System.out.print( "*" );
}
System.out.println( "" );
}
}
// Driver Code
public static void main(String[] args)
{
pattern( 7 );
}
} // This code is contributed by vt_m |
# Python3 program to print # the given pattern def pattern(n):
# This is upper half of pattern
for i in range ( 1 , n + 1 ):
for j in range ( 1 , 2 * n):
# Left part of pattern
if i > (n - j + 1 ):
print ("", end = ' ' );
else :
print ( "*" , end = '');
# Right part of pattern
if i + n - 1 > j:
print ("", end = ' ' );
else :
print ( "*" , end = '');
print ("");
# This is lower half of pattern
for i in range ( 1 , n + 1 ):
for j in range ( 1 , 2 * n):
#Left part of pattern
if i < j:
print ("", end = ' ' );
else :
print ( "*" , end = '');
# Right part of pattern
if i < 2 * n - j:
print ("", end = ' ' );
else :
print ( "*" , end = '');
print ("");
# Driver Code pattern( 7 );
# This code is contributed by mits |
// C# program to print // the given pattern using System;
class GFG
{ static void pattern( int n)
{
int i, j;
// This is upper
// half of pattern
for (i = 1; i <= n; i++)
{
for (j = 1; j <= (2 * n); j++)
{
// Left part of pattern
if (i > (n - j + 1))
Console.Write( " " );
else
Console.Write( "*" );
// Right part of pattern
if ((i + n) > j)
Console.Write( " " );
else
Console.Write( "*" );
}
Console.WriteLine( "" );
}
// This is lower
// half of pattern
for (i = 1; i <= n; i++)
{
for (j = 1; j <= (2 * n); j++)
{
// Right Part of pattern
if (i < j)
Console.Write( " " );
else
Console.Write( "*" );
// Left Part of pattern
if (i <= ((2 * n) - j))
Console.Write( " " );
else
Console.Write( "*" );
}
Console.WriteLine( "" );
}
}
// Driver Code
static public void Main ()
{
pattern(7);
}
} // This code is contributed by ajit |
<?php // PHP program to print // the given pattern function pattern( $n )
{ $i ; $j ;
// This is upper half of pattern
for ( $i = 1; $i <= $n ; $i ++)
{
for ( $j = 1; $j <= (2 * $n ); $j ++)
{
// Left part of pattern
if ( $i > ( $n - $j + 1))
echo " " ;
else
echo "*" ;
// Right part of pattern
if (( $i + $n ) > $j )
echo " " ;
else
echo "*" ;
}
printf( "\n" );
}
// This is lower half of pattern
for ( $i = 1; $i <= $n ; $i ++)
{
for ( $j = 1; $j <= (2 * $n ); $j ++)
{
// Right Part of pattern
if ( $i < $j )
echo " " ;
else
echo "*" ;
// Left Part of pattern
if ( $i <= ((2 * $n ) - $j ))
echo " " ;
else
echo "*" ;
}
echo "\n" ;
}
} // Driver Code pattern(7); // This code is contributed by m_kit ?> |
<script> // JavaScript program to print
// the given pattern
function pattern(n) {
var i, j;
// This is upper half of pattern
for (i = 1; i <= n; i++) {
for (j = 1; j <= 2 * n; j++) {
// Left part of pattern
if (i > n - j + 1)
document.write( " " );
else
document.write( "*" );
// Right part of pattern
if (i + n > j)
document.write( " " );
else
document.write( "*" );
}
document.write( "<br>" );
}
// This is lower half of pattern
for (i = 1; i <= n; i++) {
for (j = 1; j <= 2 * n; j++) {
// Right Part of pattern
if (i < j)
document.write( " " );
else
document.write( "*" );
// Left Part of pattern
if (i <= 2 * n - j)
document.write( " " );
else
document.write( "*" );
}
document.write( "<br>" );
}
}
// Driver Code
pattern(7);
</script>
|
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
Time Complexity: O(n2)
Auxiliary Space: O(1)
Program to print following pattern:
Examples :
Input : 5 Output: * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
This program is divided into four parts.
// C++ program to print the // given pattern #include <bits/stdc++.h> using namespace std;
void pattern( int n)
{ int i, j;
// This is upper half of pattern
for (i = 1; i <= n; i++)
{
for (j = 1; j <= (2 * n); j++)
{
// Left part of pattern
if (i < j)
cout << " " ;
else
cout << "*" ;
// Right part of pattern
if (i <= ((2 * n) - j))
cout << " " ;
else
cout << "*" ;
}
cout << "\n" ;
}
// This is lower half of pattern
for (i = 1; i <= n; i++)
{
for (j = 1; j <= (2 * n); j++)
{
// Left part of pattern
if (i > (n - j + 1))
cout << " " ;
else
cout << "*" ;
// Right part of pattern
if ((i + n) > j)
cout << " " ;
else
cout << "*" ;
}
cout << "\n" ;
}
} // Driver Code int main()
{ pattern(7);
return 0;
} // This code is contributed by shivanisinghss2110 |
// C program to print the // given pattern #include<stdio.h> void pattern( int n)
{ int i,j;
// This is upper half of pattern
for (i=1; i<=n; i++)
{
for (j=1; j<=(2*n); j++)
{
// Left part of pattern
if (i<j)
printf ( " " );
else
printf ( "*" );
// Right part of pattern
if (i<=((2*n)-j))
printf ( " " );
else
printf ( "*" );
}
printf ( "\n" );
}
// This is lower half of pattern
for (i=1; i<=n; i++)
{
for (j=1;j<=(2*n);j++)
{
// Left part of pattern
if (i>(n-j+1))
printf ( " " );
else
printf ( "*" );
// Right part of pattern
if ((i+n)>j)
printf ( " " );
else
printf ( "*" );
}
printf ( "\n" );
}
} // Driver Code int main()
{ pattern(7);
return 0;
} |
// Java program to print the // given pattern import java.io.*;
class GFG {
static void pattern( int n)
{
int i, j;
// This is upper half of pattern
for (i = 1 ; i <= n; i++) {
for (j = 1 ; j <= ( 2 * n); j++) {
// Left part of pattern
if (i < j)
System.out.print( " " );
else
System.out.print( "*" );
// Right part of pattern
if (i <= (( 2 * n) - j))
System.out.print( " " );
else
System.out.print( "*" );
}
System.out.println( "" );
}
// This is lower half of pattern
for (i = 1 ; i <= n; i++) {
for (j = 1 ; j <= ( 2 * n); j++) {
// Left part of pattern
if (i > (n - j + 1 ))
System.out.print( " " );
else
System.out.print( "*" );
// Right part of pattern
if ((i + n) > j)
System.out.print( " " );
else
System.out.print( "*" );
}
System.out.println( "" );
}
}
// Driver Code
public static void main(String[] args)
{
pattern( 7 );
}
} // This code is contributed by vt_m |
# Python3 program to # print the given pattern def pattern(n):
# This is upper
# half of pattern
for i in range ( 1 , n + 1 ):
for j in range ( 1 , 2 * n + 1 ):
# Left part of pattern
if (i < j):
print (" ", end = " ");
else :
print ( "*" , end = "");
# Right part of pattern
if (i < = (( 2 * n) - j)):
print (" ", end = " ");
else :
print ( "*" , end = "");
print ("");
# This is lower
# half of pattern
for i in range ( 1 , n + 1 ):
for j in range ( 1 , 2 * n + 1 ):
# Left part of pattern
if (i > (n - j + 1 )):
print (" ", end = " ");
else :
print ( "*" , end = "");
# Right part of pattern
if ((i + n) > j):
print (" ", end = " ");
else :
print ( "*" , end = "");
print ("");
# Driver Code pattern( 7 );
# This code is contributed # by mits |
// C# program to print // the given pattern using System;
class GFG
{ static void pattern( int n)
{
int i, j;
// This is upper
// half of pattern
for (i = 1; i <= n; i++)
{
for (j = 1; j <= (2 * n); j++)
{
// Left part of pattern
if (i < j)
Console.Write( " " );
else
Console.Write( "*" );
// Right part of pattern
if (i <= ((2 * n) - j))
Console.Write( " " );
else
Console.Write( "*" );
}
Console.WriteLine( "" );
}
// This is lower
// half of pattern
for (i = 1; i <= n; i++)
{
for (j = 1; j <= (2 * n); j++)
{
// Left part of pattern
if (i > (n - j + 1))
Console.Write( " " );
else
Console.Write( "*" );
// Right part of pattern
if ((i + n) > j)
Console.Write( " " );
else
Console.Write( "*" );
}
Console.WriteLine( "" );
}
}
// Driver Code
static public void Main ()
{
pattern(7);
}
} // This code is contributed by ajit |
<?php // PHP program to print // the given pattern function pattern( $n )
{ $i ; $j ;
// This is upper half
// of pattern
for ( $i = 1; $i <= $n ; $i ++)
{
for ( $j = 1; $j <= (2 * $n ); $j ++)
{
// Left part of pattern
if ( $i < $j )
echo " " ;
else
echo "*" ;
// Right part of pattern
if ( $i <= ((2 * $n ) - $j ))
echo " " ;
else
echo "*" ;
}
echo "\n" ;
}
// This is lower half of pattern
for ( $i = 1; $i <= $n ; $i ++)
{
for ( $j = 1; $j <= (2 * $n ); $j ++)
{
// Left part of pattern
if ( $i > ( $n - $j + 1))
echo " " ;
else
echo "*" ;
// Right part of pattern
if (( $i + $n ) > $j )
echo " " ;
else
echo "*" ;
}
echo "\n" ;
}
} // Driver Code pattern(7); // This code is contributed by aj_36 ?> |
<script> // Javascript program to print the // given pattern function pattern(n)
{ var i, j;
// This is upper half of pattern
for (i = 1; i <= n; i++)
{
for (j = 1; j <= (2 * n); j++)
{
// Left part of pattern
if (i < j)
document.write( " " );
else
document.write( "*" );
// Right part of pattern
if (i <= ((2 * n) - j))
document.write( " " );
else
document.write( "*" );
}
document.write( '<br>' );
}
// This is lower half of pattern
for (i = 1; i <= n; i++)
{
for (j = 1; j <= (2 * n); j++)
{
// Left part of pattern
if (i > (n - j + 1))
document.write( " " );
else
document.write( "*" );
// Right part of pattern
if ((i + n) > j)
document.write( " " );
else
document.write( "*" );
}
document.write( '<br>' );
}
} // Driver Code pattern(7); // This code is contributed by Princi Singh </script> |
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
Time Complexity: O(n2)
Auxiliary Space: O(1)
Program to print the following pattern:
Examples:
Input : 9 [For Odd number] Output: \*******/ *\*****/* **\***/** ***\*/*** ****/**** ***/*\*** **/***\** */*****\* /*******\ Input : 8 [For Even number] Output : \******/ *\****/* **\**/** ***\/*** ***/\*** **/**\** */****\* /******\
Recommended: Please try your approach on {IDE} first, before moving on to the solution.
Code implementation to print the given pattern:
// C++ program to print the given pattern #include <bits/stdc++.h> using namespace std;
void pattern( int n)
{ // for traversing of rows
for ( int i = 1; i <= n; i++) {
// for traversing of columns
for ( int j = 1; j <= n; j++) {
// conditions for left-diagonal and
// right-diagonal
if (i == j || i + j == (n + 1)) {
if (i + j == (n + 1)) {
cout << "/" ;
}
else {
cout << "\\" ;
}
}
else
cout << "*" ;
}
cout << endl;
}
} // Driver Code int main()
{ pattern(9);
return 0;
} // This code is contributed by Nitin Kumar |
// Java program to print the given pattern import java.io.*;
class pattern
{ // Function to print the given pattern
static void pattern( int n)
{
// for traversing of rows
for ( int i = 1 ; i <= n; i++)
{
// for traversing of columns
for ( int j = 1 ; j <= n; j++)
{
// conditions for left-diagonal and
// right-diagonal
if (i == j || i + j == (n + 1 )) {
if (i + j == (n + 1 )) {
System.out.print( "/" );
}
else {
System.out.print( "\\" );
}
}
else
System.out.print( "*" );
}
System.out.println();
}
}
// Driver Code
public static void main(String[] args)
{
pattern( 9 );
}
} // This code is contributed by AJAX |
# Python3 program to print the given pattern def pattern(n):
# For traversing of rows
for i in range ( 1 , n + 1 ):
# For traversing of columns
for j in range ( 1 , n + 1 ):
# Conditions for left-diagonal and right-diagonal
if i = = j or i + j = = n + 1 :
if i + j = = (n + 1 ):
print ( '/' , end = '')
else :
print ( '\\', end = ' ')
else :
print ( '*' , end = '')
print ('')
#Driver Code if __name__ = = '__main__' :
n = 8
pattern(n)
# This code is contributed by Mahendra Varma
|
// C# program to print the given pattern using System;
using System.Collections.Generic;
class GFG
{ static void pattern( int n)
{
// for traversing of rows
for ( int i = 1; i <= n; i++)
{
// for traversing of columns
for ( int j = 1; j <= n; j++)
{
// conditions for left-diagonal and
// right-diagonal
if (i == j || i + j == (n + 1)) {
if (i + j == (n + 1)) {
Console.Write( "/" );
}
else {
Console.Write( "\\" );
}
}
else
Console.Write( "*" );
}
Console.Write( "\n" );
}
}
// Driver Code
static void Main( string [] args)
{
pattern(9);
}
} |
<script> // JavaScript program to print the given pattern // Function to print the given pattern function pattern(n)
{ // for traversing of rows
for (let i = 1; i <= n; i++)
{
// for traversing of columns
for (let j = 1; j <= n; j++)
{
// conditions for left-diagonal and
// right-diagonal
if (i == j || i + j == (n + 1)) {
if (i + j == (n + 1)) {
document.write( "/" );
}
else {
document.write( "\\" );
}
}
else
document.write( "*" );
}
document.write( "<br>" );
}
} pattern(9); // This code is contributed by lokesh. </script> |
\*******/ *\*****/* **\***/** ***\*/*** ****/**** ***/*\*** **/***\** */*****\* /*******\
Time Complexity: O(n2)
Auxiliary Space: O(1)
Program to print the following pattern:
Examples :
Input : 8 Output : 7 6 5 4 3 2 1 0 6 5 4 3 2 1 0 5 4 3 2 1 0 4 3 2 1 0 3 2 1 0 2 1 0 1 0 0
Recommended: Please try your approach on {IDE} first, before moving on to the solution.
Code implementation to print the given pattern:
// C++ program to print the given pattern #include <bits/stdc++.h> using namespace std;
void pattern( int n)
{ // for traversing of rows
for ( int i = 1; i <= n; i++) {
int k = n - i;
// for traversing of columns
for ( int j = 1; j <= n; j++) {
if (j <= (n + 1) - i) {
cout << k << " " ;
k--;
}
else {
cout << " " ;
}
}
cout << endl;
}
} // Driver Code int main()
{ pattern(8);
return 0;
} // This code is contributed by Nitin Kumar |
// Java program to print the given pattern import java.util.*;
public class Main {
public static void pattern( int n) {
// for traversing of rows
for ( int i = 1 ; i <= n; i++) {
int k = n - i;
// for traversing of columns
for ( int j = 1 ; j <= n; j++) {
if (j <= (n + 1 ) - i) {
System.out.print(k + " " );
k--;
}
else {
System.out.print( " " );
}
}
System.out.println();
}
}
// Driver Code
public static void main(String args[]) {
pattern( 8 );
}
} // this code contributed by SRJ2777 |
def pattern(n):
# for traversing of rows
for i in range ( 1 , n + 1 ):
# inner loop for decrement in i values
for j in range (n - i, - 1 , - 1 ):
print (j, end = ' ' )
print ()
#Driver Code if __name__ = = '__main__' :
n = 8
pattern(n)
|
// C# program to print the given pattern using System;
using System.Collections.Generic;
class GFG
{ static void pattern( int n)
{
// for traversing of rows
for ( int i = 1; i <= n; i++) {
int k = n - i;
// for traversing of columns
for ( int j = 1; j <= n; j++) {
if (j <= (n + 1) - i) {
Console.Write(k + " " );
k--;
}
else {
Console.Write( " " );
}
}
Console.WriteLine();
}
}
// Driver Code static void Main( string [] args)
{
pattern(8);
}
} |
// JavaScript program to print the given pattern function pattern(n) {
// for traversing of rows for (let i = 1; i <= n; i++) {
let k = n - i; // for traversing of columns for (let j = 1; j <= n; j++) {
if (j <= n + 1 - i) {
console.log(k + " " );
k--; } else {
console.log( " " );
} } console.log( "<br>" );
} } // Driver Code pattern(8); |
7 6 5 4 3 2 1 0 6 5 4 3 2 1 0 5 4 3 2 1 0 4 3 2 1 0 3 2 1 0 2 1 0 1 0 0
Time Complexity: O(n2)
Auxiliary Space: O(1)
Program to print the following pattern :
Examples:
Input: 7 Output: 1 8 2 14 9 3 19 15 10 4 23 20 16 11 5 26 24 21 17 12 6 28 27 25 22 18 13 7
Recommended: Please try your approach on {IDE} first, before moving on to the solution.
Code implementation to print the given pattern:
// C++ program to print the given pattern #include <bits/stdc++.h> using namespace std;
void pattern( int n)
{ int p, k = 1;
// for traversing of rows
for ( int i = 1; i <= n; i++) {
p = k;
// for traversing of columns
for ( int j = 1; j <= i; j++) {
cout << p << " " ;
p = p - (n - i + j);
}
cout << endl;
k = k + 1 + (n - i);
}
} // Driver Code int main()
{ pattern(7);
return 0;
} // This code is contributed by Nitin Kumar |
public class Pattern {
public static void pattern( int n)
{
int p, k = 1 ;
// for traversing of rows
for ( int i = 1 ; i <= n; i++) {
p = k;
// for traversing of columns
for ( int j = 1 ; j <= i; j++) {
System.out.print(p + " " );
p = p - (n - i + j);
}
System.out.println();
k = k + 1 + (n - i);
}
}
// Driver Code
public static void main(String[] args)
{
pattern( 7 );
}
} |
# code def pattern(n):
k = 1
# for traversing of rows
for i in range ( 1 , n + 1 ):
p = k
# for traversing of columns
for j in range ( 1 ,i + 1 ):
print (p, end = ' ' )
p = p - (n - i + j)
print ()
k = k + 1 + (n - i)
#Driver Code if __name__ = = '__main__' :
n = 7
pattern(n)
|
using System;
namespace Pattern
{ class Program
{
static void Pattern( int n)
{
int p, k = 1;
// for traversing of rows
for ( int i = 1; i <= n; i++)
{
p = k;
// for traversing of columns
for ( int j = 1; j <= i; j++)
{
Console.Write(p + " " );
p = p - (n - i + j);
}
Console.WriteLine();
k = k + 1 + (n - i);
}
}
// Driver Code
static void Main( string [] args)
{
Pattern(7);
}
}
} |
<script> // JavaScript program to print the given pattern function pattern(n) {
let k = 1;
// for traversing of rows
for (let i = 1; i <= n; i++) {
let p = k;
// for traversing of columns
for (let j = 1; j <= i; j++) {
document.write(p + " " );
p = p - (n - i + j);
}
document.write( "<br>" );
k = k + 1 + (n - i);
}
} pattern(7); // This code is contributed by lokesh. </script> |
1 8 2 14 9 3 19 15 10 4 23 20 16 11 5 26 24 21 17 12 6 28 27 25 22 18 13 7
Time Complexity: O(n2)
Auxiliary Space: O(1)