Try this

1. Find the min and max in O(n) time.
2. For A.P. = mix to max/N , we find max possible subsequence.

For example
1,2,3,0,4,7,19,6,8,10,24.....(could be more but trying to show the approach)

We see min = 1 and max = 8 hence we need to try for diff = 1 to 8/size(arr)

In this case, we will find that for diff = 2, we are able to find a sequence
2,4,6,8,10 of length 5. Hence max is this sequence (Only store the first
value of every sequence and make some variable diffForMaxSequence to store
diff value 2 here)

Complexity will depend upon the range of numbers but we can look for
optimization to not go till diff = max/N

On Fri, Jul 8, 2011 at 3:18 PM, sunny agrawal <[email protected]>wrote:

> 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 <[email protected]>wrote:
>
>> Nopes....its about finding subsequence....
>>
>> On 7/8/11, rajeev bharshetty <[email protected]> wrote:
>> > Should the sequence beContinuos ???
>> >
>> > On Fri, Jul 8, 2011 at 1:18 AM, sunny agrawal
>> > <[email protected]>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
>> >> <[email protected]>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
>> >>> <[email protected]>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 <
>> [email protected]
>> >>>> > 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 <
>> [email protected]
>> >>>>> > 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 *
>> >>>>>>
>> >>>>>> --
>> >>>>>> You received this message because you are subscribed to the Google
>> >>>>>> Groups "Algorithm Geeks" group.
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>> >>>>>>
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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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>> >
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>> >
>>
>>
>> --
>> *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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-- 
Regards,
Navneet

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