When two persons have same name and we use one hashtable(HT), what happens
inside HT is entirely depends on HT implementation. In this case, HT may
chain the values when keys match. When querying the HT for this key, HT
returns all the values chained under this key and developer has to iterate
and
@Gene -
Cool algorithm! I tried before in java and messed a little to get exact
output format.
Just wondering how you came up with simple yet working code?
-Carol
On Thu, Dec 22, 2011 at 6:08 PM, Gene wrote:
> The simplest algorithm is probably to check each point against all
> others while mai
@Algoose, in worst case, this is still O(n^2), ain't it?
On Wed, Dec 21, 2011 at 12:50 PM, Algoose chase wrote:
> Find the distance between each of the points and the origin(0,0) and sort
> the points based on this distance.
> Also, divide the points based on which quadrant they belong. If the
>
