• i have used some conditions and a variable which is inc-dec .. how do you see this solution?

• int a;

• int i,j;
• scanf("%d", &a);
• int n=(a*2)-1;
• for(i=1;i<=n;i++)
• {
• for(j=1;j<=n;j++)
• {
• printf("%d ",((i-j>0&&i<(n/2)+1)||(i+j=(n/2)+1))?a--:((i-j<0 &&i>(n/2)+1)||(i+j>n+1 && i<=(n/2)+1))?++a:a);
• }
• printf("\n");
• }

At first, I created the matrix for the first quadrant and used a lot of lines to print it in order and reverse order. But later, I saw that the AI's code was much more concise. That's when I realized the pattern: the value of each element is n - d, where d is the minimum distance from the element to the matrix boundary.

Here is Printing patterns using loops solution in c - https://programmingoneonone.com/hackerrank-printing-pattern-using-loops-in-c-solution.html

include

include

include

include

int main() {

int n;
scanf("%d", &n);
// Complete the code to print the pattern.
int L = 2 * n - 1;

for (int i = 1; i <= L; i++) {
    for (int j = 1; j <= L; j++) {
        int dr = abs(i - n);
        int dc = abs(j - n);
        int d  = dr > dc ? dr : dc;
        printf("%d ", d + 1);
    }
    printf("\n");
}

return 0;

}

I have a solution which is way faster and more simple than using the array

include

include

include

include

int main() {

int n;
scanf("%d", &n);
// Complete the code to print the pattern.
for (int i = 1; i <= 2 * n - 1; i++) {
    int d = abs(i - n);
    for(int j = 1; j <= n - 1 - d; j++) printf("%d ", n - j + 1);
    for(int j = 1; j <= 2 * d + 1; j++) printf("%d ", d + 1);
    for(int j = n - 1 - d; j >= 1; j--) printf("%d ", n - j + 1);
    printf("\n");

}
return 0;

}