Hi..

you are correct ...thanks.. it can be done using DP..
and I did it :)
On Tue, Apr 3, 2012 at 1:48 AM, Carlos Guia <[email protected]>wrote:

> Do you have a constrain on k? It can be done in O(k^2).
>
> Dp(p, k) would be the count the number of arrangements of size k of the
> previous number was p.
> The transitions aren't hard, there are at most three of them, you should
> be able to figure them out.
>
> If you need more help, let me know, however, trying it by yourself is the
> way to learn it.
>
> Is k^2 is too big, well I haven't given it enough thought to know if I can
> or cannot do it better.
>
> Regards,
> Carlos Guia
> On Apr 2, 2012 1:28 PM, "vivek dhiman" <[email protected]> wrote:
>
>> + I am only looking at the count..
>>
>> for k=1  ans = 1
>> k=2 ans = 2
>> k=3 ans = 5
>> k=4 ans = 14 I guess...
>> k = 5 ans =52 I guess
>>
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-- 
Regards
Vivek Dhiman

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