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

Here, N = 5.

**Examples**:

Input Format: N = 3Result: A A B A B CInput Format: N = 6Result: A A B A B C A B C D A B C D E A B C D E F

**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.

In this pattern problem, instead of numbers, we have to print alphabets hence making the pattern look like a right-angled triangle. So, the outer loop will run for N rows and the inner loop will loop for i alphabets in each row where i is the row number. Alphabets in each row will start from A each time we enter a new row and will loop till (A+i)th alphabet in that row.

**Code**:

## C++ Code

```
#include <bits/stdc++.h>
using namespace std;
void pattern14(int N)
{
// Outer loop for the number of rows.
for(int i=0;i<N;i++){
// Inner loop will loop for i times and
// print alphabets from A to A + i.
for(char ch = 'A'; ch<='A'+i;ch++){
cout<<ch<<" ";
}
// As soon as the letters for each iteration are printed, we move to the
// next row and give a line break otherwise all letters
// 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;
pattern14(N);
return 0;
}
```

**Output **

A

A B

A B C

A B C D

A B C D E

## Java Code

```
class Main {
static void pattern14(int N)
{
// Outer loop for the number of rows.
for(int i=0;i<N;i++){
// Inner loop will loop for i times and
// print alphabets from A to A + i.
for(char ch = 'A'; ch<='A'+i;ch++){
System.out.print(ch + " ");
}
// As soon as the letters for each iteration are printed, we move to the
// next row and give a line break otherwise all letters
// 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;
pattern14(N);
}
}
```

**Output **

A

A B

A B C

A B C D

A B C D E

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]