@Sagar: So far so good, but you are not guaranteed to get an
exception. Example, int a[987] is followed in memory by char
b[10000000], which is a dictionary. You won't detect an exception
until you get to at least a[262144] (2 to the 18th). But you will pick
up plenty of garbage which may throw off your binary search.

Dave

On Aug 19, 1:26 pm, sagar pareek <[email protected]> wrote:
> Well
> sorry but i forget to mention exceptions in the solution.
> Here is the complete solution :-
>
> The key idea here is to simultaneously do a binary search
> for the end of the array as well as the key. We try to look for A[2k ] in
> the
> k-th step and catch exceptions for successive values of k till either we hit
> an exception or we hit a number greater than or equal to b. Then we do
> a binary search for b between indices 2k - 1 and 2k . The runtime of the
> search algorithm is 0 (l og 叫.
>
>
>
>
>
> On Fri, Aug 19, 2011 at 11:53 PM, Dave <[email protected]> wrote:
> > @Everyone: The problem says that the array is of UNKNOWN length, but
> > all of the solutions presented assume that the array is of INFINITE
> > length. Suppose, e.g., that the length is 987, but you don't know
> > that. Then it will be meaningless to probe at 1, 10, 100, 1000, etc,
> > or 1, 2, 4, ..., 512, 1024 because any probe beyond 987 is outside the
> > array. An address violation may occur, or arbitrary data, unrelated to
> > the data in the array may be used. I think the problem as stated is
> > unsolvable.
>
> > Dave
>
> > On Aug 19, 12:48 pm, sagar pareek <[email protected]> wrote:
> > > HI,
>
> > > I have encountered a problem :-
>
> > > You have an array of  *UNKNOWN  *length . And you have to find an element
> > > in  O(log(n)) time without using any extra space.
>
> > > --
> > > **Regards
> > > SAGAR PAREEK
> > > COMPUTER SCIENCE AND ENGINEERING
> > > NIT ALLAHABAD
>
> > --
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>
> --
> **Regards
> SAGAR PAREEK
> COMPUTER SCIENCE AND ENGINEERING
> NIT ALLAHABAD

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