@Yogesh
your solution will give maximum Contiguous AP only
it will fail for the array A[] = {1,2,3,4,5,6,8,10,12,14}
your algo will give output that there is an Longest AP of 6 elements which
is wrong

checkout this <http://theory.cs.uiuc.edu/%7Ejeffe/pubs/pdf/arith.pdf> for an
O(n^2) algorithm

On Sun, Jul 10, 2011 at 7:01 PM, Yogesh Yadav <medu...@gmail.com> wrote:

> @raj :
>
> array a[]= 2,3,5,6,7,8,10,12
> diff[]=       1,2,1,1,1,2,2
>
> maxcount=tempcount=1   // because am not takin in consideration of 0th
> index value of diff[]
>
> now in for loop
> for j=1
> check diff[j]==diff[j-1] //not equal
> so check tempcount>maxcount or not  //its also not
> so maxcount remains same and tempcount becomes 1 again
>
> now for j=2
>
> check diff[j]==diff[j-1] //not equal
> so check tempcount>maxcount or not  //its also not
> so maxcount remains same and tempcount becomes 1 again
>
> now for j=3
>
> check diff[j]==diff[j-1] // equal
> so tempcount++;
>
> and so on....
>
>
>
>
> On Sun, Jul 10, 2011 at 5:23 AM, raj singh <ankurkaku...@gmail.com> wrote:
>
>> @yogesh- can u explain with an example pls?
>>
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-- 
Sunny Aggrawal
B-Tech IV year,CSI
Indian Institute Of Technology,Roorkee

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