Inverted V pattern: Given the value of n, print the inverted V pattern.
Examples :
Input : n = 5
Output :
E
D D
C C
B B
A A
Input : n = 7
Output :
G
F F
E E
D D
C C
B B
A A
Below is the program to print the above pattern
C++
#include <bits/stdc++.h>
using namespace std;
void pattern( int n)
{
int i, j, k = 0;
for (i = n - 1; i >= 0; i--) {
for (j = n - 1; j > k; j--) {
cout << " " ;
}
cout << char (i + 65);
for (j = 1; j < (k * 2); j++)
cout << " " ;
if (i < n - 1)
cout << char (i + 65);
cout << "\n" ;
k++;
}
}
int main()
{
int n = 5;
pattern(n);
return 0;
}
|
Java
import java.io.*;
class GFG
{
static void pattern( int n)
{
int i, j, k = 0 ;
for (i = n - 1 ; i >= 0 ; i--)
{
for (j = n - 1 ; j > k; j--)
{
System.out.print( " " );
}
System.out.print(( char )(i + 65 ));
for (j = 1 ; j < (k * 2 ); j++)
System.out.print( " " );
if (i < n - 1 )
System.out.print(( char )(i + 65 ));
System.out.print( "\n" );
k++;
}
}
public static void main(String args[])
{
int n = 5 ;
pattern(n);
}
}
|
Python3
def pattern(n):
k = 0
for i in range (n - 1 , - 1 , - 1 ):
for j in range (n - 1 , k, - 1 ):
print ( ' ' , end = '')
print ( chr (i + 65 ), end = '')
for j in range ( 1 , k * 2 ):
print ( ' ' , end = '')
if i<n - 1 :
print ( chr (i + 65 ), end = '')
print ()
k + = 1
n = 5
pattern(n)
|
C#
using System;
class GFG
{
static void pattern( int n)
{
int i, j, k = 0;
for (i = n - 1; i >= 0; i--)
{
for (j = n - 1; j > k; j--)
{
Console.Write( " " );
}
Console.Write(( char )(i + 65));
for (j = 1; j < (k * 2); j++)
Console.Write( " " );
if (i < n - 1)
Console.Write(( char )(i + 65));
Console.WriteLine();
k++;
}
}
public static void Main()
{
int n = 5;
pattern(n);
}
}
|
PHP
<?php
function pattern( $n )
{
$k = 0;
for ( $i = $n - 1; $i >= 0; $i --)
{
for ( $j = $n - 1; $j > $k ; $j --)
{
echo " " ;
}
echo chr ( $i + 65);
for ( $j = 1; $j < ( $k * 2); $j ++)
echo " " ;
if ( $i < $n - 1)
echo chr ( $i + 65);
echo "\n" ;
$k ++;
}
}
$n = 5;
pattern( $n );
?>
|
Javascript
<script>
function pattern(n) {
var i,
j,
k = 0;
for (i = n - 1; i >= 0; i--) {
for (j = n - 1; j > k; j--) {
document.write( " " );
}
document.write(String.fromCharCode(i + 65));
for (j = 1; j < k * 2; j++)
document.write( " " );
if (i < n - 1)
document.write(String.fromCharCode(i + 65));
document.write( "<br>" );
k++;
}
}
var n = 5;
pattern(n);
</script>
|
Output:
E
D D
C C
B B
A A
Time Complexity: O(n2), where n represents the given input.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
V pattern: Given the value of n, print the V pattern.
Examples :
Input : n = 5
Output :
E E
D D
C C
B B
A
Input : n = 7
Output :
G G
F F
E E
D D
C C
B B
A
Below is the program to print the above pattern
C++
#include <bits/stdc++.h>
using namespace std;
void pattern( int n)
{
int i, j;
for (i = n - 1; i >= 0; i--) {
for (j = n - 1; j > i; j--) {
cout << " " ;
}
cout << char (i + 65);
for (j = 1; j < (i * 2); j++)
cout << " " ;
if (i >= 1)
cout << char (i + 65);
cout << "\n" ;
}
}
int main()
{
int n = 5;
pattern(n);
return 0;
}
|
Java
import java.io.*;
class GFG
{
static void pattern( int n)
{
int i, j;
for (i = n - 1 ; i >= 0 ; i--)
{
for (j = n - 1 ; j > i; j--)
{
System.out.print( " " );
}
System.out.print(( char )(i + 65 ));
for (j = 1 ; j < (i * 2 ); j++)
System.out.print( " " );
if (i >= 1 )
System.out.print(( char )(i + 65 ));
System.out.print( "\n" );
}
}
public static void main(String args[])
{
int n = 5 ;
pattern(n);
}
}
|
Python3
import math as mt
def pattern(n):
i = n - 1
j = 1
for i in range (n - 1 , - 1 , - 1 ):
for j in range (n - 1 , i, - 1 ):
print ( ' ' , end = '')
print ( chr (i + 65 ), end = '')
for j in range ( 1 , i * 2 ):
print ( ' ' , end = '')
if (i > = 1 ):
print ( chr (i + 65 ), end = '')
print ()
n = 5
pattern(n)
|
C#
using System;
class GFG
{
static void pattern( int n)
{
int i, j;
for (i = n - 1; i >= 0; i--)
{
for (j = n - 1; j > i; j--)
{
Console.Write( " " );
}
Console.Write(( char )(i + 65));
for (j = 1; j < (i * 2); j++)
Console.Write( " " );
if (i >= 1)
Console.Write(( char )(i + 65));
Console.WriteLine( "" );
}
}
public static void Main()
{
int n = 5;
pattern(n);
}
}
|
PHP
<?php
function pattern( $n )
{
for ( $i = $n - 1; $i >= 0; $i --)
{
for ( $j = $n - 1; $j > $i ; $j --)
{
echo " " ;
}
echo chr ( $i + 65);
for ( $j = 1; $j < ( $i * 2); $j ++)
echo " " ;
if ( $i >= 1)
echo chr ( $i + 65);
echo "\n" ;
}
}
$n = 5;
pattern( $n );
?>
|
Javascript
<script>
function pattern(n)
{
let i, j;
for (i = n - 1; i >= 0; i--)
{
for (j = n - 1; j > i; j--)
{
document.write( "  " );
}
document.write(String.fromCharCode(i + 65));
for (j = 1; j < (i * 2); j++)
document.write( "  " );
if (i >= 1)
document.write(String.fromCharCode(i + 65));
document.write( "<br>" );
}
}
let n = 5;
pattern(n);
</script>
|
Output:
E E
D D
C C
B B
A
Time Complexity: O(n2), where n represents the given input.
Auxiliary Space: O(1), no extra space is required, so it is a constant.