the reason i asked you about problem link is because the solution to your
problem depends on the way you want to use it...

1. if each time there are new N points and radius R,   and only one query
for it...then it just can't be O(n)

2. if N points are fixed and there are like 10000+ queries then you can
store it in some dataset and retrieve the result for that in O(logn)

On Fri, Jan 6, 2012 at 12:29 PM, amrit harry <[email protected]>wrote:

> i also think so.... data must be represent in some special form.... i
> heard about K-D trees but not yet studied about it....
>
>
> On Fri, Jan 6, 2012 at 12:26 PM, Lucifer <[email protected]> wrote:
>
>> @all..
>> To decide which points are within the range R, we need to look a each
>> point..Hence the lower bound would be O(N)..
>>
>> Now, for doing it in < O(N) we need to have the input being given in a
>> particular order, basically the datastructure which stores the graph
>> projects some releationship b/w all the points..
>>
>> @utkarsh..
>> Say, the graph is represented as an adjancency matrix or list with
>> edges showing the connection and weights defined the distance b/w the
>> points...and assuming that this datastructure is given to u.. then its
>> as good as applying dijikstra's to find the shortest path from A to
>> all other vertices... and check which shortest paths are smaller than
>> R...
>>
>> Basically it all depends on how the data is being represented..
>>
>> @dabbcomputer
>> Correct me if i m wrong..
>>
>> On Jan 6, 11:48 am, shady <[email protected]> wrote:
>> > how will one come to know which points are cut-out from the circle ?
>> > i fail to understand how can it be less than O(n) unless you store the
>> > given N points in a data structure ... where preprocessing itself
>> involves
>> > O(N)...
>>
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>
>
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