You have to scan every pair of points only once to get the value of 'm' and
'a', so the time complexity would be O(n^2).


On Sat, Oct 23, 2010 at 6:22 PM, Meng Yan <[email protected]> wrote:

> there are (n*(n-1))/2pairs of points. I think if we use your method, the
> time complexity should be O(n^4).
>
> Is it possible to put all points into k different domain and using
> T(n)=T(n/k)+f(n) to solve this problem?
>
>
> On Sat, Oct 23, 2010 at 7:51 PM, preetika tyagi 
> <[email protected]>wrote:
>
>> Is there any specific need to use recursion?
>>
>> One alternate is to find slope and constant (m and c) for every pair of
>> points and same value of m & c will specify the points on the same line.
>> Time complexity is O(n*n).
>>
>>
>> On Sat, Oct 23, 2010 at 4:31 PM, Meng Yan <[email protected]>wrote:
>>
>>> Given n point on the plane, find out whether any 3point on the same
>>> line.
>>>
>>> How to use recursion to solve the problem? Could you help me find the
>>> algorithm and give the time complexity?
>>>
>>> Bests,
>>> Claire
>>>
>>>
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