@DON:  DFS should not be applied because, there can exist multiple paths
from A to C, but the question is to find the shortest path. So, u might end
up getting longer path.
Otherwise you have to search for all possible paths which will be very
inefficient.


----
Saurabh Agrawal

On Tue, Sep 13, 2011 at 10:01 PM, Navneet <navneetn...@gmail.com> wrote:

> A related discussion is available here.
>
> http://stackoverflow.com/questions/1556451/how-do-sites-like-linkedin-efficiently-display-1st-2nd-3rd-level-relationship-nex
>
> On Sep 13, 8:37 pm, Don <dondod...@gmail.com> wrote:
> > I think your statement about A->B->C leads people to believe that the
> > depth is 2. If we know depth is 2, we just scan A's friends and C's
> > friends looking for a match, which is B.
> > If we don't know the length of the path, we might want to start
> > searching from A and C at the same time and look for a friend common
> > to both search trees.
> > Don
> >
> > On Sep 13, 9:57 am, JITESH KUMAR <jkhas...@gmail.com> wrote:
> >
> >
> >
> >
> >
> >
> >
> > > Depth is not mentioned, it can be any.
> > > This question was asked from me in the telephonic interview of DE Shaw.
> > > The interviewer told me reduce space complexity.
> >
> > > On Tue, Sep 13, 2011 at 6:18 PM, MeHdi KaZemI <mehdi.kaze...@gmail.com
> >wrote:
> >
> > > > I think applying BFS is good, what's the problem with space? Isn't
> the
> > > > depth gonna be at most 2 ?
> > > > If we suppose the depth is gonna be at most 2, then suppose we want
> the
> > > > path from A to C,
> > > > A has 500 friends and each of his/her friends has 500 friends too, so
> we
> > > > have to visit 500*500 nodes to find the path, am I right?
> >
> > > > On Tue, Sep 13, 2011 at 5:11 PM, Karan Thakral <
> karanthak...@gmail.com>wrote:
> >
> > > >> bfs
> >
> > > >> On Tue, Sep 13, 2011 at 5:59 PM, JITESH KUMAR <jkhas...@gmail.com>
> wrote:
> >
> > > >>> I guess you have misunderstood the problem.
> > > >>> We are not concerning about the length of path. We just have to
> find the
> > > >>> path.
> > > >>> But in the efficient way. suppose first person is having 500
> friends and
> > > >>> each of them again is having 500 friends each.
> > > >>> Applying BFS will take a lot of space.
> >
> > > >>> On Tue, Sep 13, 2011 at 5:48 PM, veera reddy <
> veeracool...@gmail.com>wrote:
> >
> > > >>>>  finding the shortest path between A and C nodes , gives required
> > > >>>> solution .
> > > >>>> We can use dijkstra's algorithm to find the shortest path ..
> >
> > > >>>> On Tue, Sep 13, 2011 at 5:43 PM, JITESH KUMAR <jkhas...@gmail.com
> >wrote:
> >
> > > >>>>> Suppose you are visiting someone's profile in fb or linkedin, you
> get
> > > >>>>> to know how you are connected to that person.
> > > >>>>>  e.g. Suppose you are visiting C's profile. you get a suggestion
> like
> > > >>>>> you are connected to him via A->B->C.
> > > >>>>> Tell efficient way to solve this problem( apart from Brute
> Force).
> >
> > > >>>>> --
> > > >>>>> *Regards
> > > >>>>> Jitesh Kumar*
> >
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