I decided that i am not going to use ai so i thinked how to solve and tried and run each time upadating code in onlinegdp compiler then i got answer after three and half hour(3:30) so it take so much time but i learn very much

include

include

include

include

int numberOfLine(int a){ if(a==1) return 1; else if(a==2) return 3; else return (a+a)-1;

}

int main() {

int n;
scanf("%d", &n);
// Complete the code to print the pattern.
int c=n,c1=n;
int num=numberOfLine(n);
for(int i=(num/2)+1;i>0;i--){ 
for(int j=n;j>=c;j--){
    printf("%d",j);}
    for(int k=num-2;k>0;k--){
        printf("%d",c1);

    }num-=2;

    for(int j=c;j<=n;j++){
        if(j!=1)
            printf("%d",j);

    }printf("\n");c1--;
    c--;

}
c=2,c1=2;
 num=numberOfLine(n);
 int temp=1;
for(int i=(num/2);i>0;i--){ 
for(int j=n;j>=c;j--){
    printf("%d",j);}
    for(int k=temp;k>0;k--){
        printf("%d",c1);

    }temp+=2;

    for(int j=c;j<=n;j++){
        if(j!=1)
            printf("%d",j);

    }printf("\n");c1++;
    c++;

}


return 0;

}

My approach is that, since the figure is symmetrical, I solve the upper-left part and then mirror it to the right, downward, and to the lower-right quadrant. Although it’s not particularly elegant.

int main() {

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

}

int main() {

int n;
int m;
int min;
int value;
scanf("%d", &n);
// Complete the code to print the pattern.

m = 2 * n -1;

for (int i = 0; i < m; i ++){
    min = 1 + abs(i-n+1);
    for (int j = 0; j < m; j ++){
        value = 1 + abs(j-n+1);
        if (value > min) printf("%d " , value);
        else printf("%d ", min);
    } 
    printf("\n");
}

return 0;

}

I tried it for more than half an hour could not solve it so turned to ChatGPT. Then i understood it. 1. First make variables to store the distance of a particular position from all the borders(top, left, right, bottom). 2. then make a new variable(I named it min_dist) to store the minimum distance from any border. 3. Make any distance from any border as min_dist. then check if the distances are lower than the min_dist. If lower then set that distance as new min_dist. 4. Lastly, print the value of that position which is (n-min_dist) HOW? See this structure as a square , the minimum distance from the outermost square to the outermost square is o ,SO,the outermost square is made of the integer n itself. and then when we go to the lower square the minimum distance from the outermost square becomes 1 ,SO,this square is made of (n-1) i.e (n-min_dist) and the same goes on for all the remaining inner squares.

I HOPE I EXPLAINED IT WELL

include

int main() {

int n;
scanf("%d", &n);

int size=(2*n-1);

for(int i=0;i<(2*n-1);i++){
    for(int j=0;j<(2*n-1);j++){
        int top=i;
        int left=j;
        int right=(size-1)-j;
        int bottom=(size-1)-i;

        int min_dist=top;
        if(bottom < min_dist){min_dist=bottom;}
        if(left < min_dist){min_dist=left;}
        if(right < min_dist){min_dist=right;}
        printf("%d\t",n-min_dist);
    }
    printf("\n");
}
return 0;

}

I modified the way to represent absolute value f(x) = |x| for easier implementation.

int main() 
{

    int n;
    scanf("%d", &n);
  	// Complete the code to print the pattern.
    int width = 2*n-1;
    int min, cur;
    for (int h = 0; h < width; h++){
        min = (n - h > 2-n + h) ? n - h : 2-n + h;
        for (int w = 0; w < width; w++){
            cur = (n - w > 2-n + w) ? n - w : 2-n + w;
            printf("%d " , (cur > min) ? cur : min);
        }
        printf("\n");
        
    }
    return 0;
}