Create a BST from given list such that team that has won should be
right child of node and and teams that has lost should be left child
of node and then traverse the tree in reverse in order.
Karan
On Thu, May 23, 2013 at 10:58 PM, bharat b bagana.bharatku...@gmail.com wrote:
@Don : you are
This is not necessarily possible.
If you have teams A, B, and C, it is possible that
A beat B
B beat C
C beat A.
Based on the first two games the ranking should be A,B,C. But based on
the third game, C should be ranked higher than A.
Don
On May 23, 11:06 am, Nishant Pandey
@DON,
For example if in a particular order, the teams appeared as T1, T2, T3, T4
… then team T1 had lost to T2, T2 had lost to T3, and T3 had lost to T4… It
may be possible that T3 lost to T1 .. but that need not be taken into
consideration while writing the order. Only the neighboring elements
http://www.careercup.com/question?id=14804702
On Thu, May 23, 2013 at 8:53 PM, Nishant Pandey
nishant.bits.me...@gmail.com wrote:
@DON,
For example if in a particular order, the teams appeared as T1, T2, T3, T4
… then team T1 had lost to T2, T2 had lost to T3, and T3 had lost to T4… It
may
If you create a directed graph where each node is a team and an edge
exists from A-B if A lost to B, then find a Hamiltonian Path in the
graph. That path will be the sequence you need.
Don
On May 23, 12:02 pm, bharat b bagana.bharatku...@gmail.com wrote:
@Don : you are correct but hamiltanion path problem is NPC right?
quicksort algorithm is good solution. I already shared algorithm in above
link.
On Thu, May 23, 2013 at 10:25 PM, Don dondod...@gmail.com wrote:
If you create a directed graph where each node is a team and an edge
exists from
