@Sanjay
yeah its the very basic idea that comes in mind
but is your index searching log n ?
i think no !!
if yes then tell me how?

On Fri, Aug 19, 2011 at 11:24 PM, Sanjay Rajpal <srn...@gmail.com> wrote:

> I forgot to mention one thing, at each comparison, store the index at which
> we searched previously.
>
> Sanju
> :)
>
>
>
> On Fri, Aug 19, 2011 at 10:53 AM, Sanjay Rajpal <srn...@gmail.com> wrote:
>
>> 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 <sagarpar...@gmail.com>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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>>
>>
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-- 
**Regards
SAGAR PAREEK
COMPUTER SCIENCE AND ENGINEERING
NIT ALLAHABAD

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