# 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.

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<stdio.h>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;}`

### One comment on “Print Right Diamond Number Pattern Type4”

• 1MV20CS110Somen

#include
int main()
{
int rows,i,j=0,k=0,x=0;
printf(“enter the number of rows\n”);
scanf(“%d”,&rows);
for(i=1;i<=rows;i++)
{
j=0;
k=0;
if(i<=(rows+1)/2)
{
while(j(rows+1)/2)
{
while(k<(rows-i+1))
{
printf("%d",i-x);
k++;
}
x=x+2;
printf("\n");
}
}
return 0;
} 1