Here is one Brute Force solution O(n^3)
for each i and j in array(j > i) {where a[i] is first term of AP and a[j] is
second term}
compute d = a[j]-a[i]
and now from j+1 to end of the array search for a+2d,a+3d,a+4d
................
and keep track of longest :)
On Fri, Jul 8, 2011 at 1:28 AM, Piyush Sinha <ecstasy.piy...@gmail.com>wrote:
> Nopes....its about finding subsequence....
>
> On 7/8/11, rajeev bharshetty <rajeevr...@gmail.com> wrote:
> > Should the sequence beContinuos ???
> >
> > On Fri, Jul 8, 2011 at 1:18 AM, sunny agrawal
> > <sunny816.i...@gmail.com>wrote:
> >
> >> @rajiv
> >> if Count = 2 means 3 elements isn't it a,a+d,a+2d
> >> else according to you
> >> for case 10 12 14 24 26 28
> >> diff 2 2 10 2 2
> >> diff 2 has count 4 so will you say ap of 4 elements with diff 2
> >>
> >> On Fri, Jul 8, 2011 at 1:06 AM, rajeev bharshetty
> >> <rajeevr...@gmail.com>wrote:
> >>
> >>> @sunny Keep count of longest repeated element in diff i.e 2 so count =2
> >>> so
> >>> ap of 2 elem with diff 2 .
> >>>
> >>> On Fri, Jul 8, 2011 at 1:03 AM, sunny agrawal
> >>> <sunny816.i...@gmail.com>wrote:
> >>>
> >>>> @rajiv
> >>>> Fails i think
> >>>> think for 10 12 24 26
> >>>> diff is 2 12 2
> >>>> so do you want to say there is an AP pf 3 elements with d = 2, i can't
> >>>> see any :P
> >>>> your solution fails because there can be many APs in the array with
> the
> >>>> same value of d and you will finish up by combining all those APs
> >>>>
> >>>>
> >>>> On Fri, Jul 8, 2011 at 12:55 AM, rajeev bharshetty <
> rajeevr...@gmail.com
> >>>> > wrote:
> >>>>
> >>>>>
> >>>>> Check the differences between the adjacent elements and store the
> >>>>> differenecs in diff[i] array
> >>>>> then scan through the array .
> >>>>> then keep a count for all the repeated diff elements ,the sequence of
> >>>>> indexes with max count is the solution .
> >>>>>
> >>>>>
> >>>>>
> >>>>>
> >>>>> On Thu, Jul 7, 2011 at 11:43 PM, Piyush Sinha <
> ecstasy.piy...@gmail.com
> >>>>> > wrote:
> >>>>>
> >>>>>> Given an array of integers A, give an algorithm to find the longest
> >>>>>> Arithmetic progression in it, i.e find a sequence i1 < i2 < … < ik,
> >>>>>> such that
> >>>>>>
> >>>>>> A[i1], A[i2], …, A[ik] forms an arithmetic progression, and k is the
> >>>>>> largest possible.
> >>>>>>
> >>>>>> The sequence S1, S2, …, Sk is called an arithmetic progression if
> >>>>>>
> >>>>>> Sj+1 – Sj is a constant.
> >>>>>>
> >>>>>> --
> >>>>>> *Piyush Sinha*
> >>>>>> *IIIT, Allahabad*
> >>>>>> *+91-8792136657*
> >>>>>> *+91-7483122727*
> >>>>>> *https://www.facebook.com/profile.php?id=100000655377926 *
> >>>>>>
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> >>>> --
> >>>> Sunny Aggrawal
> >>>> B-Tech IV year,CSI
> >>>> Indian Institute Of Technology,Roorkee
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> >>
> >>
> >>
> >> --
> >> Sunny Aggrawal
> >> B-Tech IV year,CSI
> >> Indian Institute Of Technology,Roorkee
> >>
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> --
> *Piyush Sinha*
> *IIIT, Allahabad*
> *+91-8792136657*
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>
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--
Sunny Aggrawal
B-Tech IV year,CSI
Indian Institute Of Technology,Roorkee
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