simple and easy answer

for (int i = 0; i < size; i++) {
        for (int j = 0; j < size; j++) {
            int min = (i<j?i:j)<(size-i-1<size-j-1?size-i-1:size-j-1)?(i<j?i:j):(size-i-1<size-j-1?size-i-1:size-j-1);
            printf("%d ", n - min);
        }
        printf("\n");

takes the minimum out of i,j,size-i-1 and size-j-1, where size is 2n-1

core component of this code: for(int x=0;x=n;x--) { for(int y=2*(n-1)-z;y>=z;y--){ a[x][y]=n-z; a[y][x]=n-z; } z++;}

include

void main( ) { int n; printf("Give input"); scanf("%i",&n); for(int i=1; i <= 2*n-1;i++) { int k=1; for(int j=1;j <= 2*n-1; j++) { for(int m=1;mi) k --; if(j>=2*n-i+1) k --; break; } printf("%i ",n-k+1); } printf("\n"); } }

include

include

include

include

int main() {

int n;
scanf("%d", &n);
// Complete the code to print the pattern.
int arr_siz = ((n-1)*2) + 1;
int **arr = (int**)malloc(arr_siz * sizeof(int*));
for (int i = 0; i < arr_siz; i++)
    arr[i] = (int*)malloc(arr_siz * sizeof(int));

int arr_loop = (arr_siz/2) + 1;
int num_dec = 0;
for(int i =0; i < arr_loop; i++)
{
    int  dec_val = 0;
    num_dec = i;
    for(int j=0; j< arr_loop; j++)
    {
        arr[i][j] = n-dec_val;
        arr[i][arr_siz -1 - j] = n-dec_val;
        if(num_dec)
        {
            dec_val += 1;
            num_dec--;
        }
    }
    memcpy(arr[arr_siz-1-i], arr[i],arr_siz * sizeof(int));
}
for (int i = 0; i < arr_siz; i++)
{
    for (int j = 0; j < arr_siz; j++)
    {
        printf("%d ", arr[i][j]);
    }
    printf("\n");
}       

/* Code for further processing and free the
   dynamically allocated memory */

for (int i = 0; i < arr_siz; i++)
    free(arr[i]);
return 0;

}

include

include

include

include

int main() {

int n;
scanf("%d", &n);
int min=0;
for(int i=1;i<=2*n-1;i++){
    for(int j=1;j<=2*n-1;j++){
        int a = i;
        if(i>n) a = 2*n - i;
        int b = j;
        if(b>n) b = 2*n - j;
        if(a<b) min = a;
        else min = b;
        printf("%d ",n+1-min);
    }
    printf("\n");
}
// Complete the code to print the pattern.
return 0;

}