@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 *
>>>>
>>>> --
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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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