This might not lead to a correct solution always. Say, if B and C were not
friends, you would get two paths which are of same length. But both B and C
are still mutual friends of A and D. Besides, I am not sure if taking
A-B-C-D as a path would make a good representation.
@Shashank: If you are
Regarding the explanation of the suffix tree method, here is the link:
http://algo2006.csie.dyu.edu.tw/paper/2/A24.pdf
On Tue, Jan 24, 2012 at 8:05 PM, Ashish Goel ashg...@gmail.com wrote:
http://www.akalin.cx/longest-palindrome-linear-time
lovely explanation..
Best Regards
Ashish Goel
Then I will go with Atul's solution. BFS should work. Consider the graph
with nodes representing as people and edges to be the friendship
association. For any i,j pair you need to look for the paths of length 2
from one node to another. For finding it out for all the pairs, it will
have to be done
I am still not clear on the question!
Do it suggest that we have m people with n friends each and the problem is
to find mutual friends b/w any two people?
And by the mutual friend, does it mean that he has to be a friend of both
of the people or can he be indirectly linked? Your previous comment
