Need help in Google Kickstart Round E problem B

1

I have written the code in java but i think logic is clear.
i tried to first add no of buildings A, then added no of buildings seen niether by A or B. then c buildings of height n and then B buildings. I have taken care of overlap of c a and b.
if someone could help with test cases or logic

    import java.util.Scanner;
    class Solution
    {
        public static void main(String args[])
        {
               int t,n,a1,b,c,w,k,i,j,v;
          Scanner sc= new Scanner (System.in);
            t=sc.nextInt();
          w=t;
        while(t>0)
       {
           t--;
           n=sc.nextInt();
           a1=sc.nextInt();
           b=sc.nextInt();
           c=sc.nextInt();
           k=n;v=0;
           long a[] =new long[n];
           System.out.print("Case #"+(w-t)+": ");
           if((a1+b+1)!=(n+c))
            {System.out.print("IMPOSSIBLE");
            System.out.println();}
            else
            {
           i=0;
           for(i=0;i<(a1-c);i++)
           {
               a[i]=2;
           }
           for(j=0;j<(n+c-(a1+b)) && (a1-c)>0;j++)
            {
                a[i]=1;
                i++;
                v++;
            }
           
           for(j=1;j<=c;j++)
            {
                a[i]=n;
                i++;
            }
            for(j=0;j<(n+c-(a1+b)-v) && (b-c)>0;j++)
            {
                a[i]=1;
                i++;
            }
           for(i=i;i<n;i++)
            a[i]=n-2;
           
            for(i=0;i<n;i++)
             System.out.print(a[i]+" ");
             System.out.println();
            }
       }
       }
    }
2

please format your code

1 Like
3

Why this condition? Check the case
1
5 2 2 1
Your code gives the answer IMPOSSIBLE
But there exist answers like 4 5 1 1 4

My approch was to first find the number of buildings seen only by A, only by B, by both and by none. Let’s say these are a1, b1, c, and d = n-(a1+b1+c). If d is negative, the answer is impossible otherwise I assigned (n-1) a1 times, n c times and (n-1) b1 times. Then d 1s can be inserted between any two buildings.

3 Likes
4

okay now i see. will update the code and try again.

5

done. sorry for the late reply