CareerCup

i think it can be done like this cuboid all sides are equal so u can simply have int point instead int point[3] and based on size of the point we can store it in list .eg the input node are like this 50 30 49 20 11 2 22 then in first node store 50 in first list next 30 comes so again move 50 to second node and store 30 in first next 49 comes now move 50 to 3rd position 49 in 2nd now 20 move 49 to 3rd and 50 4th and 20 to 2nd then for 11 make it as first then for 2 make this as first node now for 22 all the entry in first node is done so check 22 comes in between which node 20 and 30 so put 22 as first entry on node 20 if u do like this searching will be more easy then bst only thing here is considering each node size is unique .please let me know any thing wrong.

@Anonymous :- From where did you get this idea "in cuboid all sides are equal". I learnt this property for cubes only. Anyway, I asked interviewer if co-ordinates are given or not. He told me nothing is given. If co-ordinates along with length,breadth and hight of cuboid would have been given then no need to use 6 pointers here, we can traverse simply by comparing co-ordinates moving accordingly.

Comments please.

I think he was not satisfied because every node can be reached many different ways. Suppose if you reached a node traversing n nodes then it can be reached 6^n ways. This is exponential and therefore BFS would be inefficient. I would say go to the first node by traversing as much left, below and behind then traverse all the nodes layer by layer, row by row, cell by cell.

I thought it as graph problem. Can you explain a bit more ?

i think take a 3D array to store each cuboid and take another 3D array to store chain of cuboid like
struct cuboid
{
int east;
int west;
int north;
int south;
int top;
int bottom;
}
cuboid chain_of_cuboid[][][][][][];

in above its not 3D array its 6D array

above both comment is not valid ignore it

In one of my comment, I have written that I asked interviewer that whether co-ordinates of points are given or not. He said NO. So we can't use this approach.

Oct tree could have been used in case these were cubes (but why complicate stuff, its overkill) and k-d tree cannot be, they have Nearest Neighbour algo (NN) but k-d trees have left child smaller and right side larger property which doesn't fit here.