**Problem Statement:** Given an integer **N, **print the following pattern :

Here, N = 5.

**Examples**:

Input Format: N = 3Result: * * ** ** ****** ** ** * *Input Format: N = 6Result: * * ** ** *** *** **** **** ***** ***** ************ ***** ***** **** **** *** *** ** ** * *

**Solution**

** Disclaimer**:

*Don’t jump directly to the solution, try it out yourself first.*

**Approach**:

There are 4 general rules for solving a pattern-based question :

- We always use nested loops for printing the patterns. For the outer loop, we count the number of lines/rows and loop for them.
- Next, for the inner loop, we focus on the number of columns and somehow connect them to the rows by forming a logic such that for each row we get the required number of columns to be printed.
- We print the numbers inside the inner loop.
- Observe symmetry in the pattern or check if a pattern is a combination of two or more similar patterns or not.

In this problem, we have to print a butterfly-like star pattern. This pattern is very similar to Pattern 10 in this series as here we can see that for Row 1 we have 2 stars, and 8 spaces, and for Row 2 we have 4 stars and 6 spaces, and so on. Also, after the nth row, the stars decrease and the spaces increase. So, the outer loop will run for 2*n -1 times ( n times when spaces > stars and then n-1 when stars > spaces ). There will be 3 inner loops, one to print stars, then spaces, and then again stars. Spaces will decrement by 2 as i increases and when i reaches n, then spaces decrement by 2 every time we enter a new row. When i<n, the stars printed in each row are 2*i, and as soon as i>n, the stars in each row would be twice of 2*n-i.

**Code**:

## C++ Code

```
#include <bits/stdc++.h>
using namespace std;
void pattern20(int n)
{
// initialising the spaces.
int spaces = 2*n-2;
// Outer loop for printing row.
for(int i = 1;i<=2*n-1;i++){
// stars for first half
int stars = i;
// stars for the second half.
if(i>n) stars = 2*n - i;
//for printing the stars
for(int j=1;j<=stars;j++){
cout<<"*";
}
//for printing the spaces
for(int j = 1;j<=spaces;j++){
cout<<" ";
}
//for printing the stars
for(int j = 1;j<=stars;j++){
cout<<"*";
}
// As soon as the stars for each iteration are printed, we move to the
// next row and give a line break otherwise all stars
// would get printed in 1 line.
cout<<endl;
if(i<n) spaces -=2;
else spaces +=2;
}
}
int main()
{
// Here, we have taken the value of N as 5.
// We can also take input from the user.
int N = 5;
pattern20(N);
return 0;
}
```

**Output **

* *

** **

*** ***

**** ****

**********

**** ****

*** ***

** **

* *

## Java Code

```
class Main {
static void pattern20(int n)
{
// initialising the spaces.
int spaces = 2*n-2;
// Outer loop for printing row.
for(int i = 1;i<=2*n-1;i++){
// stars for first half
int stars = i;
// stars for the second half.
if(i>n) stars = 2*n - i;
//for printing the stars
for(int j=1;j<=stars;j++){
System.out.print("*");
}
//for printing the spaces
for(int j = 1;j<=spaces;j++){
System.out.print(" ");
}
//for printing the stars
for(int j = 1;j<=stars;j++){
System.out.print("*");
}
// As soon as the stars for each iteration are printed, we move to the
// next row and give a line break otherwise all stars
// would get printed in 1 line.
System.out.println();
if(i<n) spaces -=2;
else spaces +=2;
}
}
public static void main(String[] args) {
// Here, we have taken the value of N as 5.
// We can also take input from the user.
int N = 5;
pattern20(N);
}
}
```

**Output **

* *

** **

*** ***

**** ****

**********

**** ****

*** ***

** **

* *

Special thanks toplease check out this article.Priyanshi Goelfor contributing to this article on takeUforward. If you also wish to share your knowledge with the takeUforward fam,If you want to suggest any improvement/correction in this article please mail us at [email protected]