# Print Right Diamond Number Pattern Type4

3

44

555

6666

555

44

3

## 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 pyramid. 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. Inside this loop print i+3.
5. Else run a different loop from j=i to jThe loop should be structured as for( j=i ; j
6. Inside this loop print (r-i+2).
7. 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()
{//declaring integer variables i,j for loops and r for number of rows
int i,j,r; //Asking user for input
printf("Enter the number of rows(odd) :\n");
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 digits per each row
{
printf("%d",i+3);   //printing numbers
}
}
else                            //else condition to print the bottom half
{
for(j=i;j<r;j++)          //loop for printing numbers
{
printf("%d",r-i+2); //printing numbers
}
}

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

## DISPLAYING OUTPUT: If you want to do the same code but by doing 2 inputs then the code will change as –

Input:
3  4
Output:
3
44
555
6666
555
44
3

Input :
4  4
Output:
4
55
666
7777
666
55
4

### Code 1

`#include int main(){  int i,j,s,N,count=0;  scanf(“%d%d”,&s,&N);    for(i=s;count<4;count++)    {      for(j=0;j<count+1;j++)        printf(“%d”,i);      printf(“\n”);      i=i+1;    }    for(i=s+N-2;count>0;count–)    {      for(j=0;j<count-1;j++)        printf(“%d”,i);      printf(“\n”);      i=i-1;     }  return 0;}`