int main() {

int n;
scanf("%d", &n);
int a[10000],m=((2*n)-1);
int p=0,q=m,r=n; 
// Complete the code to print the pattern.
for(int o=0;o<r;o++){
for(int i=p;i<q;i++){
    a[i]=n;
    }  
 for(int j=0;j<m;j++){
    printf("%d ",a[j]);
     }    
 p++;
 q--;
 n--;
 printf("\n");
}
n=n+2;
for(int o=0;o<r-1;o++){
 p--;
 q++;
 for(int i=p;i<q;i++){
    a[i]=n;
    }  
 for(int j=0;j<m;j++){
    printf("%d ",a[j]);
     }    

 n++;
 printf("\n");
}


return 0;

}

• 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;

}