i dont understand it..if k is at position an let suppose....so to
check at that position dont you have to traverse till that position
...is thr anything else than the head of list...??or i understood
wrongly...??

On Mon, Jul 18, 2011 at 9:55 PM, sagar pareek <sagarpar...@gmail.com> wrote:
> Update on 2nd line
>
> 2.    if( ptr2=ptr1->next->next.......(5 or 10 times) ) goto 3.       else
> make linear search till NULL encounter and exit with the solution
>
> On Mon, Jul 18, 2011 at 7:41 PM, sagar pareek <sagarpar...@gmail.com> wrote:
>>
>> i have one approach :-
>>
>> first compare root->data  and k
>> if k is very much greater than root->data then set next=5or10 ur choice
>>
>> else set next 2or3  ur choice
>> take two pointers ptr1 and ptr2
>>
>> now lets take k is much greater than root->data
>> then
>> 1. set ptr1=root //initially
>> 2. if( ptr2=ptr1->next->next.......(5 or 10 times) ) else make linear
>> search till NULL encounter
>> 3. if ptr2->data==k return its position
>> 4. else if (ptr2->data>k) set ptr1=ptr2 goto 2
>> 5. else traverse the ptr1 upto ptr2, if found return its position else
>> return fail
>>
>> if anyone has more efficient solution then pls tell  :)
>> On Mon, Jul 18, 2011 at 6:53 PM, Dumanshu <duman...@gmail.com> wrote:
>>>
>>> You have a sorted linked list of integers of some length you don't
>>> know and it keeps on increasing. You are given a number k. Print the
>>> position of the number k.
>>> Basically, you have to search for number k in the ever growing sorted
>>> list and print its position.
>>>
>>> Please write the complexity of whatever solution you propose.
>>>
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>>
>>
>>
>> --
>> Regards
>> SAGAR PAREEK
>> COMPUTER SCIENCE AND ENGINEERING
>> NIT ALLAHABAD
>>
>
>
>
> --
> Regards
> SAGAR PAREEK
> COMPUTER SCIENCE AND ENGINEERING
> NIT ALLAHABAD
>
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