# C++ Program to print pattern 2*N Number of rows

## To Print 2*N Number of rows pattern

In this article we will learn to print 2*N row patterns of numbers, where we need to use many loops in C++ programming language.Here ‘i’ loop is used to access each line from 1 to N and ‘j’ loop is used to print values in each line.
Here we are given you solution to deal with this problem.

## Implementation:-

• Given input from user i.e number of lines required (2*N value).
• Take a result variable (say ‘num’).
• Take two loops and two subloops one for each line (say ‘i’) and other for each digit in a particular line (say ‘j’).
• Here ‘i’ loop is used to access each line from 1 to N and ‘j’ loop is used to print values in each line.
• Decrement the ‘a’ value at the beginning of each line to 2*N values.
• Repeat the ‘i’ loop until it reaches ‘N’ lines.
`//C++ Program to print 2*N Number of rows pattern#include <iostream>using namespace std;int main(){    int n=4,num=n-1;   // Predefined input    for(int i=1;i<=n;i++)    {        for(int j=1;j<=i;j++)             cout<<num;        num++;        cout<<endl;    }        num--;    for(int i=n;i>=1;i--)    {        for(int j=1;j<=i;j++)            cout<<num;        num--;        cout<<endl;    }    return 0;}`
`class Main {    public static void main(String[] args)    {        int n=4,num=n-1;        for(int i=1;i<=n;i++)        {            for(int j=1;j<=i;j++)                System.out.print(num);            num++;            System.out.println();        }                num--;        for(int i=n;i>=1;i--)        {            for(int j=1;j<=i;j++)                System.out.print(num);            num--;            System.out.println();        }    }}`

### One comment on “C++ Program to print pattern 2*N Number of rows”

• CHINTA

n = int(input())
m=n-1
for i in range(1,2*n+1):
if i < n or i == n:
print(i*str(m))
m=m+1
else:
m = m-1
print((2*n-i)*str(m-1))