That would take all the fun away....what if you are given only the address
of the array?This wont work in that case

On Mon, Aug 22, 2011 at 10:39 PM, asdqwe <ayushgoel...@gmail.com> wrote:

> If i am not wrong, the only possible solution can be
> len=sizeof(arr)/sizeof(arr[0])
> i.e. find the length from the array itself.
>
>
> On Aug 22, 9:01 pm, saurabh singh <saurab...@gmail.com> wrote:
> > @dave or anyone??????? response please
> >
> >
> >
> >
> >
> >
> >
> >
> >
> > On Sun, Aug 21, 2011 at 12:43 PM, saurabh singh <saurab...@gmail.com>
> wrote:
> > > kkk...not sure
> > > assume no number is greater than 1000(I mentioned There has to be some
> > > additional constraints to make the problem solvable)....
> > > Now check 1st element if not the desired element keep multiplying with
> 2
> > > the previous range till either one of these condition is satisfied
> > > *1.An exception is caught*
> > > *2.Number greater than 1000 occurs.*
> > > suppose this happens for *1024 *for the given example.
> > > then we will check out for (512+1024)/2 th element for the above
> condition.
> > > If true than again branch like binary search.This way can element which
> on
> > > left side doesn't gives any exception and maintains the constraints
> while on
> > > the right it violates the same.So we may land up with the desired index
> and
> > > can then perform binary search.......
> >
> > > PS:There are lots of assumption in this approach and the more I write
> the
> > > more I get convinced that its a plain stupid idea...
> >
> > > --
> > > Saurabh Singh
> > > B.Tech (Computer Science)
> > > MNNIT ALLAHABAD
> >
> > --
> > Saurabh Singh
> > B.Tech (Computer Science)
> > MNNIT ALLAHABAD
>
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-- 
Saurabh Singh
B.Tech (Computer Science)
MNNIT ALLAHABAD

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