Print a shape with * using C

Trying to learn how to code with C.
I’m trying to print a draw according to the amount of lines the user asks. This for an school project.
For example, this if the user asks for a drawing of 7 lines.

****
 ***
  **
   *
   **
   ***
   ****

Or this if the user asks for a 5 lines drawing.

***
 **
  *   
  **
  ***

With some help from Chat GPT (a lot) I’ve got this so far:

#include <stdio.h>

int main() {
    int n, i, j, k;

    // Ask the user for a number of lines.
    printf("Insert the numbers of lines to form the draw: ");
    scanf("%d", &n);

    // Validate if the number of lines requested by the user is odd.
    if (n % 2 == 0) {
        printf("The number of lines must be an odd amount.\n");
        return 1; 
    }

    // Print upper boddy of the draw.
    for (i = 0; i < n/2+1; i++) {
        for (j = 0; j < i; j++) {
            printf(" ");
        }
        for (k = 0; k < n/2+1-i; k++) {
            printf("*");
        }
        printf("\n");
    }

    // Print lower body of the draw.
    for (i = n/2-1; i >= 0; i--) {
        for (j = 0; j < i; j++) {
            printf(" ");
        }
        for (k = 0; k < n/2+1-i; k++) {
            printf("*");
        }
        printf("\n");
    }

    return 0; 
}

Problem I get with this is I get something like this instead:

*****
 ****
  ***
   **
    *
   **
  ***
 ****
*****

What do I have to modify?

>Solution :

Essentially the posted code, prints the upper part properly. So let us not disturb it. We need to work only on the lower part of printing.

Here as well, your *s are printed properly. So you would need to think about only the print of spaces which is done by the j loop. Basically, you need to print n/2 (constant) number of spaces for this offset.

// Print lower body of the draw.
for (i = n/2-1; i >= 0; i--) {
    for (j = 0; j < n/2; j++) { // updated this line
        printf(" ");
    }
    for (k = 0; k < n/2+1-i; k++) {
        printf("*");
    }
    printf("\n");
}