Program to print number with star pattern
We have to print the pattern as given in the below example.
Examples :
Input : 5
Output :
1
1*2
1*2*3
1*2
1
Input : 9
Output :
1
1*2
1*2*3
1*2*3*4
1*2*3*4*5
1*2*3*4
1*2*3
1*2
1
C++
#include <iostream>
using namespace std;
void display( int n)
{
int sp = n / 2, st = 1;
for ( int i = 1; i <= n; i++) {
for ( int j = 1; j <= sp; j++) {
cout << " " ;
}
int count = 1;
for ( int k = 1; k <= st; k++) {
if (k % 2 == 0)
cout << "*" ;
else
cout << count++;
}
cout << "\n" ;
if (i <= n / 2) {
sp = sp - 1;
st = st + 2;
}
else {
sp = sp + 1;
st = st - 2;
}
}
}
int main()
{
int n = 5;
display(n);
return 0;
}
|
Java
import java.util.Scanner;
class Pattern
{
void display( int n)
{
int sp = n / 2 , st = 1 ;
for ( int i = 1 ; i <= n; i++)
{
for ( int j = 1 ; j <= sp; j++)
{
System.out.print( " " );
}
int count = 1 ;
for ( int k = 1 ; k <= st; k++)
{
if (k % 2 == 0 )
System.out.print( "*" );
else
System.out.print(count++);
}
System.out.println();
if (i <= n / 2 )
{
sp = sp - 1 ;
st = st + 2 ;
}
else
{
sp = sp + 1 ;
st = st - 2 ;
}
}
}
public static void main(String[] args)
{
int n = 5 ;
Pattern p = new Pattern();
p.display(n);
}
}
|
Python3
def display(n):
sp = n / / 2
st = 1
for i in range ( 1 , n + 1 ):
for j in range ( 1 , sp + 1 ):
print (end = " " )
count = 1
for k in range ( 1 , st + 1 ):
if (k % 2 = = 0 ):
print ( "*" , end = "")
else :
print (count, end = "")
count + = 1
print ()
if (i < = n / / 2 ):
sp = sp - 1
st = st + 2
else :
sp = sp + 1
st = st - 2
n = 5
display(n)
|
C#
using System;
class Pattern
{
void display( int n)
{
int sp = n / 2, st = 1;
for ( int i = 1; i <= n; i++)
{
for ( int j = 1; j <= sp; j++)
{
Console.Write( " " );
}
int count = 1;
for ( int k = 1; k <= st; k++)
{
if (k % 2 == 0)
Console.Write( "*" );
else
Console.Write(count++);
}
Console.WriteLine();
if (i <= n / 2)
{
sp = sp - 1;
st = st + 2;
}
else
{
sp = sp + 1;
st = st - 2;
}
}
}
public static void Main()
{
int n = 5;
Pattern p = new Pattern();
p.display(n);
}
}
|
PHP
<?php
function display( $n )
{
$sp = $n / 2;
$st = 1;
for ( $i = 1; $i <= $n ; $i ++)
{
for ( $j = 1; $j <= $sp ; $j ++)
{
echo " " ;
}
$count = 1;
for ( $k = 1; $k <= $st ; $k ++)
{
if ( $k % 2 == 0)
echo "*" ;
else
echo $count ++;
}
echo "\n" ;
if ( $i <= $n / 2)
{
$sp = $sp - 1;
$st = $st + 2;
}
else
{
$sp = $sp + 1;
$st = $st - 2;
}
}
}
$n = 5;
display( $n );
?>
|
Javascript
<script>
function display(n)
{
var sp = n / 2,
st = 1;
for ( var i = 1; i <= n; i++) {
for ( var j = 1; j <= sp; j++) {
document.write( " " );
}
var count = 1;
for ( var k = 1; k <= st; k++) {
if (k % 2 == 0)
document.write( "*" );
else
document.write(count++);
}
document.write( "<br>" );
if (i <= n / 2)
{
sp = sp - 1;
st = st + 2;
} else
{
sp = sp + 1;
st = st - 2;
}
}
}
var n = 5;
display(n);
</script>
|
Output:
1
1*2
1*2*3
1*2
1
Time Complexity: O(n2), where n represents the given input.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Last Updated :
17 Feb, 2023
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