# Print Number Star Right Diamond Pattern Type1

1

2*2

3*3*3

4*4*4*4

3*3*3

2*2

1

## PREREQUISITE:

Basic knowledge of C language and use of loops.

## ALGORITHM:

1. Take the number of rows as input from the user and store it in any variable.(‘r‘ in this case).
2. Run a loop ‘r’ number of times to iterate through each of the rows. From i=0 to i<r. The loop should be structured as for( i=0 ; i
3. Use an if condition to to print the top half of the triangle. if (i<=r/2). Then run a loop from j=0 to j<=i. The loop should be structured as for(j=0 ; j<=i ; j++)
4. Run a nested if statement if(j!=0)  then print star and i+1.
5. Else print only i+1.
6. Else statement for the outer if statement: run a loop from j=i to jThe loop should be structured as for(j=i ; j<r; j++)
7. Inside this loop run an if statement if(j!=i) then print star and r-i.
8. Else just print r-i.
9. Inside the main loop print a newline to move to the next line after each row is printed.

## CODE IN C:

### Code

```#include
int main()
{
int i,j,r; //declaring integer variables i,j for loops and r for number of rows
printf("Enter the number of rows(odd) :\n"); //Asking user for input
scanf("%d",&r); //taking number of rows and saving it in variable r
for(i=0;i<r;i++) // loop for number of rows
{
if(i<=(r/2)) //if condition to print the top half
{
for(j=0;j<=i;j++) // loop for stars per each row
{
if(j!=0)
{
printf("*%d",i+1); //printing stars
}
else
{
printf("%d",i+1); //printing stars
}
}
}
else //else condition to print the bottom half
{
for(j=i;j<r;j++) //loop for printing
{
if(j!=i)
{
printf("*%d",r-i); //printing stars
}
else
{
printf("%d",r-i); //printing stars
}
}
}

printf("\n"); // printing newline after each row
}
}```

## DISPLAYING OUTPUT:

However if you had to minor changes and print the following pattern then code will change as –

Input :
3
Output:
1
2*2
3*3*3
3*3*3
2*2
1

Input :
4
Output:
1
2*2
3*3*3
4*4*4*4
4*4*4*4
3*3*3
2*2
1

### Code

`#include int main(){    int i,j,k,N,count=0;    scanf(“%d”,&N);    for(i=1;i<=N;i++)    {        k=1;        for(j=0;j<i;j++)        {            printf(“%d”,i);            if(k<i)            {                printf(“*”);                k=k+1;            }        }        printf(“\n”);    }    for(i=N;i>0;i–)    {        k=1;        for(j=0;j<i;j++)        {            printf(“%d”,i);            if(k<i)            {                printf(“*”);                k=k+1;            }        }        printf(“\n”);    }    return 0;}`