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 :(
>
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yes :(
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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.
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