@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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