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

Here, N = 5.

**Examples**:

Input Format: N = 3Result: 1 1 12 21 123321Input Format: N = 6Result: 1 1 12 21 12 321 1234 4321 12345 54321 123456654321

**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 want to print a combination of a numbered pyramid and a reverse-numbered pyramid. So, as per our observation in each row, numbers are printed from 1 to the row number and then some spaces and then again numbers from 1 to the row number but in reverse order. So, the outer loop will run from 1 to N and there will be three inner loops for numbers, spaces, and then again numbers.

The first inner loop will have numbers printed from 1 to the row number, the second will print the spaces ( 8 spaces in row 1, 6 spaces in row 2, and so on) and then the last loop will run from row number to 1 in decreasing manner. For spaces, we can say that initially, spaces are 2*(N-1) for Row 1 where N is the total no. of rows and then the spaces decrease by 2 in each iteration till the last row is reached.

**Code**:

## C++ Code

```
#include <bits/stdc++.h>
using namespace std;
void pattern12(int N)
{
// initial no. of spaces in row 1.
int spaces = 2*(N-1);
// Outer loop for the number of rows.
for(int i=1;i<=N;i++){
// for printing numbers in each row
for(int j=1;j<=i;j++){
cout<<j;
}
// for printing spaces in each row
for(int j = 1;j<=spaces;j++){
cout<<" ";
}
// for printing numbers in each row
for(int j=i;j>=1;j--){
cout<<j;
}
// As soon as the numbers for each iteration are printed, we move to the
// next row and give a line break otherwise all numbers
// would get printed in 1 line.
cout<<endl;
// After each iteration nos. increase by 2, thus
// spaces will decrement by 2.
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;
pattern12(N);
return 0;
}
```

**Output **

1 1

12 21

123 321

1234 4321

1234554321

## Java Code

```
class Main {
static void pattern12(int N)
{
// initial no. of spaces in row 1.
int spaces = 2*(N-1);
// Outer loop for the number of rows.
for(int i=1;i<=N;i++){
// for printing numbers in each row
for(int j=1;j<=i;j++){
System.out.print(j);
}
// for printing spaces in each row
for(int j = 1;j<=spaces;j++){
System.out.print(" ");
}
// for printing numbers in each row
for(int j=i;j>=1;j--){
System.out.print(j);
}
// As soon as the numbers for each iteration are printed, we move to the
// next row and give a line break otherwise all numbers
// would get printed in 1 line.
System.out.println();
// After each iteration nos. increase by 2, thus
// spaces will decrement by 2.
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;
pattern12(N);
}
}
```

**Output **

1 1

12 21

123 321

1234 4321

1234554321

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]