Step 1: first make a diff[][]

Step 2: search the diff in matrix
Complexity will be O(n^2)


On Sun, Jul 10, 2011 at 6:24 AM, JAIDEV YADAV <jaid...@gmail.com> wrote:

> try Matrix search...
>
>
> On Sun, Jul 10, 2011 at 7:34 PM, Yogesh Yadav <medu...@gmail.com> wrote:
>
>> yeah then it will be possible in O(n^2)
>>
>>
>> On Sun, Jul 10, 2011 at 5:57 AM, Yogesh Yadav <medu...@gmail.com> wrote:
>>
>>> @sunny : thanks ..i got it...
>>>
>>>
>>> On Sun, Jul 10, 2011 at 5:55 AM, sunny agrawal 
>>> <sunny816.i...@gmail.com>wrote:
>>>
>>>> Longest AP for that Example is of 7 elements........
>>>> 2,4,6,8,10,12,14
>>>> now see your mistake .......
>>>> as i have already told you that u r looking for only contiguous AP's and
>>>> that won't work
>>>>
>>>>
>>>> On Sun, Jul 10, 2011 at 7:18 PM, Yogesh Yadav <medu...@gmail.com>wrote:
>>>>
>>>>> @sunny: my algo will give 6 as answer ....and i guess its right... if
>>>>> am wrong plz explain where and why my logic is wrong
>>>>>
>>>>> On Sun, Jul 10, 2011 at 5:37 AM, sunny agrawal <
>>>>> sunny816.i...@gmail.com> wrote:
>>>>>
>>>>>> @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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>>>>>
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>>>>
>>>>
>>>>
>>>> --
>>>> Sunny Aggrawal
>>>> B-Tech IV year,CSI
>>>> Indian Institute Of Technology,Roorkee
>>>>
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>>>
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>
>
>
> --
> JaiDev Yadav
> (National Yoga Champion)
> Computer Engg. Dept.
> National Institute of Technology
> Kurukshetra,Haryana
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