Print Right Diamond Number Pattern Type7

PRINTING PATTERN:

2

33

444

5555

5555

444

33

2

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) 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 count.
5. Outside this loop increment count and print a newline
6. Else decrement count
7. Run a different loop from j=0 to j<r-i+1. The loop should be structured as for( j=0 ; j<r-i+1 ; j++).
8. Inside this loop print count.
9. After the loop print a newline.

CODE IN C:

#include
int main()
{
int i,j,r,count; //declaring integer variables i,j for loops , r for number of rows
printf("Enter the number of rows/columns :\n"); //asking user for the number of rows;
scanf("%d",&r); //taking number of rows and saving in variable r
count=2; //intialising count =3
for(i=1;i<=r;i++) //loop for number of rows
  {
    if(i<=r/2)
      {
        for(j=1;j<=i;j++) //loop to print digit in every column of a row
          {
            printf("%d",count); //printing digit
          }
        count++; //incrementing count
        printf("\n"); //printing newline

      }
    else
      {
        count--;
        for(j=0;j<r-i+1;j++)
          {

            printf("%d",count);
          }

        printf("\n");
     }
  }
}

TAKING INPUT:

DISPLAYING OUTPUT:

If you want to do the same code, but by using 2 inputs one for initializing the first term and then the number of times you want it to be printed.

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

Program:
#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;
}