with matrix this problem will be in O(n^2).
We don' have to just check the max number present in matrix. We have to
check as
array a[]= 1,2,3,4,5,6,8,10,12,14
diff[][]= 1 2 3 4 5 6 8 10 12 14
2 0 1 2 3 4 6 8 10 12
3 -1 0 1 2 3 5 7 9 11
4 -2 -1 0 1 2 4 6 8 10
5 0
6 0
8 0
10 0
12 0
14 0
now in this diff[][] you have to search for if the diff[2][3] =1 then there
should be diff[3][i]=1 and then diff[i][j]=1 ...and so on ...this will give
n A.P of 6 elements
but check for diff[2][4]=2 then diff[4][6]=2 ....and so on...this will give
an A.P of 7
and its solvable in O(n^2)
On Mon, Jul 11, 2011 at 5:01 PM, sunny agrawal <sunny816.i...@gmail.com>wrote:
> @Divye sir
> yeah that will work fine if D is in reasonable limits ......
>
>
> On Mon, Jul 11, 2011 at 4:26 PM, DK <divyekap...@gmail.com> wrote:
>
>> @Ritu: Your solution is incorrect.
>>
>> Consider
>>
>> 1 3 41 43 47 49 90 100 110
>>
>> Maximum repeated 'd' value:
>> '2' for the pairs (1,3), (41,43), (47,49) = 3 repeats. However, the
>> sequences themselves are not part of an AP.
>> The longest AP is of size 3 - (90, 100, 110) with a d of 10.
>>
>>
>> --
>> DK
>>
>> http://twitter.com/divyekapoor
>> http://www.divye.in
>>
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>
>
> --
> Sunny Aggrawal
> B-Tech IV year,CSI
> Indian Institute Of Technology,Roorkee
>
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