Programs to print Interesting Patterns
Last Updated :
02 May, 2025
Program to print the following pattern:
Examples :
Input : 5
Output:
* * * * * * * * * *
* * * * * * * *
* * * * * *
* * * *
* *
* *
* * * *
* * * * * *
* * * * * * * *
* * * * * * * * * *
This program is divided into four parts.
C++
// 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
<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>
<?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
?>
Output* * * * * * * * * * * * * *
* * * * * * * * * * * *
* * * * * * * * * *
* * * * * * * *
* * * * * *
* * * *
* *
* *
* * * *
* * * * * *
* * * * * * * *
* * * * * * * * * *
* * * * * * * * * * * *
* * * * * * * * * * * * * *
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++
// 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
<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>
<?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
?>
Output* *
* * * *
* * * * * *
* * * * * * * *
* * * * * * * * * *
* * * * * * * * * * * *
* * * * * * * * * * * * * *
* * * * * * * * * * * * * *
* * * * * * * * * * * *
* * * * * * * * * *
* * * * * * * *
* * * * * *
* * * *
* *
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 :
\******/
*\****/*
**\**/**
***\/***
***/\***
**/**\**
*/****\*
/******\
Code implementation to print the given pattern:
C++
// 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>
Output\*******/
*\*****/*
**\***/**
***\*/***
****/****
***/*\***
**/***\**
*/*****\*
/*******\
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
Code implementation to print the given pattern:
C++
// 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);
Output7 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
Code implementation to print the given pattern:
C++
// C++ program to print the given pattern
#include <bits/stdc++.h>
using namespace std;
void pattern(int n)F
{
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>
Output1
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)
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