U can put them into a BST and then root will dominate the left sub tree , i
think that must of lesser time complexity then O(n^2)
On Wed, Aug 1, 2012 at 5:54 PM, Sambhavna Singh wrote:
> yes :(
>
> --
> You received this message because you are subscribed to the Google Groups
> "Algorithm Geeks"
yes :(
--
You received this message because you are subscribed to the Google Groups
"Algorithm Geeks" group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group
i can think of O(n^2) approach .. i guess you are expecting with less time
complexity
On Wed, Aug 1, 2012 at 9:49 AM, Sambhavna Singh wrote:
> If we are given a number of points on the XY-plane,
> [(x0,y0),(x1,y1),(x2,y2),...].
>
> A point (xi,yi) is dominant to another point (xj,yj) iff xi>xj an
If we are given a number of points on the XY-plane,
[(x0,y0),(x1,y1),(x2,y2),...].
A point (xi,yi) is dominant to another point (xj,yj) iff xi>xj and yi>yj.
Calculate all pairs of points such that one dominates the other.
--
You received this message because you are subscribed to the Google Gro