You can do it very easily.
I assume array is sorted and contains integers.
Say start at position 1, if value at that index is equal to the value to be
found, return index.
else if value at that index is greater than the value to be found, we got an
interval to search in.
else(value at that index is smaller than the value to be found)
search at location 10,then 100, then 1000 till you find an interval.
Once you find an interval, perform Binary Search on this and get element in
O(log n).
Got it ?
Sanju
:)
On Fri, Aug 19, 2011 at 10:48 AM, 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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