# Pattern – 3: Right-Angled Number Pyramid

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

Here, N = 5.

Examples:

```Input Format: N = 3
Result:
1
1 2
1 2 3

Input Format: N = 6
Result:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
1 2 3 4 5 6
```

### 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 upright, so the inner loop will run for the row number in each iteration. For eg: 1 number for row 1, 5 numbers for row 5, and so on. The only difference between this pattern and pattern 2 is that here we print numbers looping from 1 to the row number for each row instead of printing stars.

Code

## C++ Code

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

void pattern3(int N)
{
// This is the outer loop which will loop for the rows.
for (int i = 1; i <= N; i++)
{
// This is the inner loop which loops for the columns
// no. of columns = row number for each line here.
// Here, we print numbers from 1 to the row number
// instead of stars in each row.
for (int j = 1; j <=i; j++)
{
cout <<j<<" ";
}

// As soon as 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;
}
}

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

pattern3(N);

return 0;
}
``````

Output

1
1 2
1 2 3
1 2 3 4
1 2 3 4 5

## Java Code

``````class Main {
static void pattern3(int N)
{
// This is the outer loop which will loop for the rows.
for (int i = 1; i <= N; i++)
{
// This is the inner loop which loops for the columns
// no. of columns = row number for each line here.
// Here, we print numbers from 1 to the row number
// instead of stars in each row.
for (int j = 1; j <= i; j++)
{
System.out.print(j+" ");
}

// As soon as 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();
}
}

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;
pattern3(N);
}
}
``````

Output

1
1 2
1 2 3
1 2 3 4
1 2 3 4 5