@icy
It's still there except that you'll get a different question.
That page promises you a telephone interview if you solve the challenge
but I don't know how true that is for non-US guys ..
i solved one question two weeks back  .. and no one contacted me till now ..


~raju

On Tue, Oct 25, 2011 at 3:27 AM, icy` <[email protected]> wrote:

> is this contest still going? if so, where ?  i have a solution that
> does
> (100, 1267650600228229401496703205376 )    (just one hundred 1's)
> in 0.03 seconds in an older ruby on an older pc
>
> I'd like to submit ;P
>
>
> On Oct 21, 10:48 pm, sunny agrawal <[email protected]> wrote:
> > yea i know 1st Approach is much better and is Only O(N^2) for
> > precomputing all the values for nck and then O(k) for finding no of
> > bits set in The Kth number and another loop of O(k) to find the
> > required number
> >
> > i posted 2nd approach in the context to vandana's tree approach of
> > sorting 2^N numbers, rather simply sort the numbers in the array...
> > and this approach is O(N*2^N)
> >
> > On 10/21/11, sravanreddy001 <[email protected]> wrote:
> >
> >
> >
> >
> >
> >
> >
> >
> >
> > > @Sunny.. why do we need an O(2^N) complexity?
> >
> > > for a value of N=40-50, the solution is not useful..
> >
> > > but, your 1st approach is lot better and i have got it too..
> >
> > > 1. O(N) complexity to search the k. (k bits in the numbers)  x- (sigma
> 1->k
> > > (n C i))
> > > 2. again, keep substracting (k-i) for i= 0->k-1  so.. O(k) here
> > > and recursively performing step 2. (worst case complexity is O(T))
> > > where T = nCk
> >
> > > O(N) + O(T) ==> O(T) as it dominates the given number. unless it
> doesn't
> > > fall in the range.. or   equivalently -->  max( O(T), O(N) )
> >
> > > --
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> >
> > --
> > Sunny Aggrawal
> > B.Tech. V year,CSI
> > Indian Institute Of Technology,Roorkee
>
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