Program 333: Number Diamond Pattern 2 for given pattern

Program 333: Number Diamond Pattern 2 for given pattern
Method I:

//    11
//   1221
//  123321
// 12344321
//1234554321
// 12344321
//  123321
//   1221
//    11
#include<stdio.h>
main()
{
  int i,j,k,count,number;
  printf("Enter number of rows\n");
  scanf("%d",&number);
  count=number-1;
  for(i=1;i<=number;i++)
  {
    for(k=1;k<=count;k++)
    {
      printf(" ");
    }
    count--;
    for(j=1;j<=i;j++)
    {
      printf("%d",j);
    }
    for(j=i;j>=1;j--){
     printf("%d",j);
    }
    printf("\n");
  }
  count=1;
  for(i=number-1;i>=1;i--)
  {
    if(i!=(2*(number)-1))
    {
     for(k=1;k<=count;k++)
     {
      printf(" ");
     }
     count++;
     for(j=1;j<=i;j++)
     {
      printf("%d",j);
     }
     for(j=i;j>=1;j--){
     printf("%d",j);
     }
     printf("\n");
  }
 }
}

Method II:

#include<stdio.h>
main()
{
 int i,j,rows,k,space,num=1;
 printf("Enter number of rows\n");
 scanf("%d",&rows);
    space=rows-1;
 for(i=1;i<=2*rows-1;i++)
 {
  for(j=1;j<=space;j++)
  printf(" ");
  for(k=1;k<=num;k++)
  printf("%d",k);
  for(k=num;k>=1;k--)
  printf("%d",k);
  if(i<=rows-1)
  {
    num++;
    space--;
  }
  else
  {
   space++; 
   num--;
  }
  printf("\n");
 }
}

Explanation:

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Output: