# Pattern-5: Inverted Right Pyramid

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

Here, N = 5.

Examples:

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

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

### 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 ‘*’ 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 pattern, we run the outer loop for N times as we have to print N rows and since we have to print a right-angled triangle/pyramid which must be inverted, the inner loop will run in decreasing order of stars, for eg: Row 1 (i=0) would contain N stars, Row 2 (i=1) would contain (N -1) stars and so on.

Code

## C++ Code

``````#include <bits/stdc++.h>
using namespace std;

void pattern5(int N)
{
// This is the outer loop which will loop for the rows.
for (int i = 0; i < N; i++)
{
// This is the inner loop which loops for the columns
// no. of columns = (N - row index) for each line here.
for (int j =N; j>i; j--)
{
cout <<"* ";
}

// As soon as 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;
}
}

int main()
{
// Here, we have taken the value of N as 5.
// We can also take input from the user.
int N = 5;

pattern5(N);

return 0;
}
``````

Output

* * * * *
* * * *
* * *
* *

## Java Code

``````class Main {
static void pattern5(int N)
{
// This is the outer loop which will loop for the rows.
for (int i = 0; i < N; i++)
{
// This is the inner loop which loops for the columns
// no. of columns = (N - row index) for each line here.
for (int j = N; j > i; j--)
{
System.out.print("* ");
}

// As soon as 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();
}
}

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;
pattern5(N);
}
}
``````

Output

* * * * *
* * * *
* * *
* *