How can i print the pattern given below (rhombus shape) in C? - c

Question

Expected output:
I have already printed the rhombus shape but I cant get it to print this number pattern (1 232 34543...)
Here is my code.
If anyone can help me I would appreciate it very much.
Thanks in advance.
int main()
{
int i, j, rows;
int stars, spaces;
printf("Enter rows to print : ");
scanf("%d", &rows);
int a;
stars = 1;
spaces = rows - 1;
/* Iterate through rows */
for(i=1; i<rows*2; i++)
{
/* Print spaces */
for(j=1; j<=spaces; j++)
printf(" ");
/* Print stars */
for(j=1; j<stars*2; j++){
printf("*");
}
/* Move to next line */
printf("\n");
if(i<rows)
{
spaces--;
stars++;
}
else
{
spaces++;
stars--;
}
}
getch();
return 0;
}

Answer 1

I did not put my answer to let you figure it out, but since there is already one I show how I did it too. It seems to work. (Only the changed part).
for(j=1; j<stars*2; j++){
if (j < stars)
if (i < rows) // upper left triangle
printf("%c", '0' + ((j + i - 1) % 10));
else // lower left triangle
printf("%c", '0' + ((j + rows * 2 - i - 1) % 10));
else
if (i < rows) // upper right triangle
printf("%c", '0' + (2 * stars - j + i - 1) % 10);
else // lower right triangle
printf("%c", '0' + (2 * stars - j + rows * 2 - i - 1) % 10);
}
Making a bit more compact and getting rid of the "character mentallity" in the printf.
for(char c, j = 1; j < stars * 2; j++) {
if (j < stars)
c = i < rows ? (j + i - 1) : (j + rows * 2 - i - 1);
else
c = i < rows ? (2 * stars - j + i - 1) : (2 * stars - j + rows * 2 - i - 1);
printf("%d", c % 10);
}

Answer 2

Actually there are only two parts that need to be changed. (Since other people have posted answers)
Note: the following solution requires some knowledge on the absolute value function.
First, before the forloop, you declare another variable which stores the value of rows
int totalRows = rows;
then, inside the forloop, change your printf("*"); to the following:
printf("%d",(stars - abs(j-stars) -1 + totalRows - abs(totalRows - i)) %10);
Let me break it down into two sections: row wise and column wise.
Row wise:
When we look at each row, we can see that the number is increasing starting at index 0 until it reaches the center (at index stars), then it starts to decrease. This is the shape of an inverted absolute value function shifted upward. The -1 is to account for starting at index 0 instead of index 1.
stars - abs(j-stars) -1
prints a shape of
This accounts for the increments and decrements in the rows
Column wise:
When we look at each column, we see a similar pattern: the number increases starting at index 1 until center, then it starts to decrease. Another shape of inverted absolute function shifted upward.
row - abs(row - i)
This accounts for the increments and decrements in the columns
When you combine the two together, and take mod 10 (which only keeps the last digit). You get the desired output
printf("%d",(stars - abs(j-stars) -1 + totalRows - abs(totalRows - i)) %10);
//totalRows has the initial value of rows

Answer 3

I found one solution to your problem as below...
#include <stdio.h>
int main()
{
int i, j, k, x, rows, half_row=1;
int stars, spaces;
printf("Enter rows to print : ");
scanf("%d", &rows);
int a;
stars = 1;
spaces = rows - 1;
/* Iterate through rows */
for(i=1; i<rows*2; i++)
{
/* Print spaces */
for(j=1; j<=spaces; j++)
printf(" ");
/* Print numbers */
if(rows >= i){ /* Print First Half */
x = i;
for(k=0;k<i;k++){
if(x>=10){
x=x%10;
}
printf("%d", x);
x++;
}
x--;
for(k=1;k<i;k++){
x--;
printf("%d", x);
if(x==0){
x=10;
}
}
}
else{ /* Print Second Half */
x--;
for(k=1;k<=i-(2*half_row);k++){
if(x==10){
x=0;
}
printf("%d", x);
x++;
}
x--;
for(k=1;k<=i-(2*half_row)-1;k++){
x--;
printf("%d", x);
if(x==0){
x=10;
}
}
half_row++;
}
/* Move to next line */
printf("\n");
if(i<rows)
{
spaces--;
stars++;
}
else
{
spaces++;
stars--;
}
}
getch();
return 0;
}