@Bittu...When we choose low as 2^(n-1) and high as 2^n we are reducing the
complexity from O(n) (Linear Search ) to logn (base2) . Here the thing is to
apply normal binary search between low and high  and thats where we decrease
the complexity . If the required element is not in this range we change
low=high and high =  2*high and again apply Binary Search. In the code
before applying binary search u each time check whether k <a[high] . If not
we change low and high else apply binary search here . Ideally the
complexity would be lot less than log(n) but since the no is infinite and k
can also be taken very very high then say k lies between 2*(10^9) and
4*(10^9) which is a very high number in Itself  . n is that very high
number


This approach wont work if the infinite array is not sorted

Regards
Ankur

On Mon, Oct 24, 2011 at 7:09 PM, Bittu Sarkar <bittu...@gmail.com> wrote:

> @Ankur Don't think there's any major reason for using powers of 2 except
> the fact that computing the powers of 2 can be done very efficiently than
> computing the powers of any other number. Complexity in any case remains the
> same.
>
>
> On 24 October 2011 10:29, rahul sharma <rahul23111...@gmail.com> wrote:
>
>> +1 ankur
>>
>>
>> On Mon, Oct 24, 2011 at 1:26 AM, Ankur Garg <ankurga...@gmail.com> wrote:
>>
>>> Use Binary Search
>>>
>>> start = 2^n-1 high =2^n where n=0,1....
>>>
>>> On Mon, Oct 24, 2011 at 12:28 AM, sunny agrawal <sunny816.i...@gmail.com
>>> > wrote:
>>>
>>>> hint 1: try to find 2 indexes i, j such that a[i] <= K <= a[j]
>>>>
>>>>
>>>> On Sun, Oct 23, 2011 at 11:23 PM, Ankuj Gupta <ankuj2...@gmail.com>wrote:
>>>>
>>>>> Given a sorted array of Infinite size, find an element ‘K’ in the
>>>>> array without using extra memory in O (lgn) time
>>>>>
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>>>> Sunny Aggrawal
>>>> B.Tech. V year,CSI
>>>> Indian Institute Of Technology,Roorkee
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> Bittu Sarkar
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> Department of Computer Science & Engineering
> Indian Institute of Technology Kharagpur
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