@Preetika: Then you have to look for duplicates in an array of n(n-1)/
2 real numbers. I think this takes the complexity above O(n^2).

Dave

On Oct 23, 10:54 pm, preetika tyagi <[email protected]> wrote:
> 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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