Program to print Even Odd Number Pyramid
Last Updated :
06 Sep, 2022
Given the total number of rows as n, the task is to print the given pattern.
*
1*
*2*
1*3*
*2*4*
1*3*5*
*2*4*6*
1*3*5*7*
*2*4*6*8*
1*3*5*7*9*
.
.
Examples:
Input: n = 5
Output:
*
1*
*2*
1*3*
*2*4*
Input: n = 10
Output:
*
1*
*2*
1*3*
*2*4*
1*3*5*
*2*4*6*
1*3*5*7*
*2*4*6*8*
1*3*5*7*9*
Below is the solution to the above problem:
C++
#include <iostream>
using namespace std;
void Pattern( int n)
{
int i, j, k;
for (i = 1; i <= n; i++) {
for (j = 1, k = i; j <= i; j++, k--) {
if (k % 2 == 0) {
cout << j;
}
else {
cout << "*" ;
}
}
cout << "\n" ;
}
}
int main()
{
int n = 5;
Pattern(n);
return 0;
}
|
C
#include <stdio.h>
void Pattern( int n)
{
int i, j, k;
for (i = 1; i <= n; i++) {
for (j = 1, k = i; j <= i; j++, k--) {
if (k % 2 == 0) {
printf ( "%d" , j);
}
else {
printf ( "*" );
}
}
printf ( "\n" );
}
}
int main()
{
int n = 5;
Pattern(n);
return 0;
}
|
Java
import java.util.Scanner;
class Pattern {
static void display( int n)
{
int i, j, k;
for (i = 1 ; i <= n; i++) {
for (j = 1 , k = i; j <= i; j++, k--) {
if (k % 2 == 0 ) {
System.out.print(j);
}
else {
System.out.print( "*" );
}
}
System.out.print( "\n" );
}
}
public static void main(String[] args)
{
int n = 5 ;
display(n);
}
}
|
Python3
def display(n):
for i in range ( 1 , n + 1 ):
k = i
for j in range ( 1 , i + 1 ):
if k % 2 = = 0 :
print (j, end = '')
else :
print ( '*' , end = '')
k - = 1
print ()
n = 5
display(n)
|
C#
using System;
class GFG
{
static void display( int n)
{
int i, j, k;
for (i = 1; i <= n; i++)
{
for (j = 1, k = i; j <= i; j++, k--)
{
if (k % 2 == 0)
{
Console.Write(j);
}
else
{
Console.Write( "*" );
}
}
Console.Write( "\n" );
}
}
public static void Main()
{
int n = 5;
display(n);
}
}
|
PHP
<?php
function display( $n )
{
$i ;
$j ;
$k ;
for ( $i =1; $i <= $n ; $i ++)
{
for ( $j =1, $k = $i ; $j <= $i ; $j ++, $k --)
{
if ( $k %2==0){
echo $j ;
}
else {
echo "*" ;
}
}
echo "\n" ;
}
}
$n = 5;
display( $n );
?>
|
Javascript
<script>
function Pattern(n)
{
var i, j, k;
for (i = 1; i <= n; i++) {
for (j = 1, k = i; j <= i; j++, k--) {
if (k % 2 == 0) {
document.write( j);
}
else {
document.write( "*" );
}
}
document.write( "<br>" );
}
}
var n = 5;
Pattern(n);
</script>
|
Output
*
1*
*2*
1*3*
*2*4*
Complexity Analysis:
- Time Complexity: O(n2), where n represents the given input.
- Auxiliary Space: O(1), no extra space is required, so it is a constant.
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