@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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