Given the value of n, print the pattern.
Examples :
Input : n = 5
Output :
0
101
21012
3210123
432101234
Input : n = 7
Output :
0
101
21012
3210123
432101234
54321012345
6543210123456
Below is the program to print the above pattern
C++
#include <bits/stdc++.h>
using namespace std;
void print( int n)
{
int var1 = 1, var2 = 1;
for ( int i = 0; i < n; i++) {
for ( int j = n - 1; j > i; j--) {
cout << " " ;
}
for ( int k = 1; k <= var1; k++) {
cout << abs (k - var2);
}
var1 += 2;
var2++;
cout << "\n" ;
}
}
int main()
{
int n = 5;
print(n);
return 0;
}
|
Java
class GFG
{
static void print( int n)
{
int var1 = 1 , var2 = 1 ;
for ( int i = 0 ; i < n; i++)
{
for ( int j = n - 1 ; j > i; j--)
{
System.out.print( " " );
}
for ( int k = 1 ; k <= var1; k++)
{
System.out.print(Math.abs(k - var2));
}
var1 += 2 ;
var2++;
System.out.println();
}
}
public static void main (String[] args)
{
int n = 5 ;
print(n);
}
}
|
Python3
def printt(n):
var1 = 1
var2 = 1
for i in range (n):
for j in range (n - 1 ,i, - 1 ):
print ( " " , end = "")
for k in range ( 1 , var1 + 1 ):
print ( abs (k - var2), end = "")
var1 + = 2
var2 + = 1
print ()
n = 5
printt(n)
|
C#
using System;
class GFG
{
static void print( int n)
{
int var1 = 1, var2 = 1;
for ( int i = 0; i < n; i++)
{
for ( int j = n - 1; j > i; j--)
{
Console.Write( " " );
}
for ( int k = 1; k <= var1; k++)
{
Console.Write(Math.Abs(k - var2));
}
var1 += 2;
var2++;
Console.WriteLine();
}
}
public static void Main ()
{
int n = 5;
print(n);
}
}
|
PHP
<?php
function print1( $n )
{
$var1 = 1;
$var2 = 1;
for ( $i = 0; $i < $n ; $i ++)
{
for ( $j = $n - 1; $j > $i ; $j --)
{
echo " " ;
}
for ( $k = 1; $k <= $var1 ; $k ++)
{
echo abs ( $k - $var2 );
}
$var1 += 2;
$var2 ++;
echo "\n" ;
}
}
$n = 5;
print1( $n );
?>
|
Javascript
<script>
function print( n)
{
let var1 = 1, var2 = 1;
for (let i = 0; i < n; i++)
{
for (let j = n - 1; j > i; j--)
{
document.write( " " );
}
for (let k = 1; k <= var1; k++) {
document.write(Math.abs(k - var2));
}
var1 += 2;
var2++;
document.write( "<br/>" );
}
}
let n = 5;
print(n);
</script>
|
Output :
0
101
21012
3210123
432101234
Time Complexity : O(n*n) ,where n is the number of rows in pattern
Auxiliary Space : O(1)