:)

On Fri, Aug 19, 2011 at 11:43 PM, Sanjay Rajpal <srn...@gmail.com> wrote:

> Thats wat I said, it depends. Searching in the interval will compensate
> reaching the index earlier.
> So both are almost equivalent.
>
> Sanju
> :)
>
>
>
> On Fri, Aug 19, 2011 at 11:12 AM, sagar pareek <sagarpar...@gmail.com>wrote:
>
>> Well i think it depends...
>> because range of   x and 10x is more than  i and 2i
>> no doubt multiple of 10 will give us early index but then to find number
>> in b/w indexes is more than of 2^i
>>
>>
>> On Fri, Aug 19, 2011 at 11:38 PM, Sanjay Rajpal <srn...@gmail.com> wrote:
>>
>>> Multiplication by 10 or 2^i , it depends.
>>>
>>> Multiplication by 10 will be faster, I think.
>>>
>>> Sanju
>>> :)
>>>
>>>
>>>
>>> On Fri, Aug 19, 2011 at 11:05 AM, sagar pareek <sagarpar...@gmail.com>wrote:
>>>
>>>> hmmm ok
>>>> i found a solution in which index searching is done by 2^i
>>>> which is more optimal
>>>> multiplication by 10 or 2 power i ??    i=0,1,2,3.....
>>>>
>>>> On Fri, Aug 19, 2011 at 11:30 PM, Sanjay Rajpal <srn...@gmail.com>wrote:
>>>>
>>>>> See at each step you are multiplying the index to be compared by
>>>>> 10(say), this increase is exponential.
>>>>> Therefore the search is exponential and complexity is log n. Base
>>>>> depends on the factor by which you are multiplying for the next index to 
>>>>> be
>>>>> compared.
>>>>>
>>>>> Sanju
>>>>> :)
>>>>>
>>>>>
>>>>>
>>>>> On Fri, Aug 19, 2011 at 10:57 AM, sagar pareek 
>>>>> <sagarpar...@gmail.com>wrote:
>>>>>
>>>>>> @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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>>>>>> COMPUTER SCIENCE AND ENGINEERING
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>>>> COMPUTER SCIENCE AND ENGINEERING
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>> 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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