Here is the idea I was thinking of:
- start from the root: if(root->data < k) move to position 2 in the
list.
- in this way every time move the pointer to the position double the
current position and compare th eelement of node with k( moving here
is form 1 - 2 , 2-4, 4-8 ,8-16 ......)
- suppose you found a certain node at position n whose node->data >
k , then now u only have to search for k between index n/2 to n (i.e.
if you found 16th element >k then search for k between 8th and 16ht
element)..
correct me if any flaws.....
On Jul 19, 3:38 am, Dumanshu <duman...@gmail.com> wrote:
> @Gaurav: Ever Increasing means that you don't know the length of the
> list. So you have to assume some index n, traverse the list upto that
> point and check the results. If not found increment the n to some
> higher value.
> We are basically trying to improve the complexity by taking higher and
> higher jumps for n.
>
> On Jul 19, 2:03 am, Gaurav Popli <abeygau...@gmail.com> wrote:
>
>
>
>
>
>
>
> > yeah...im saying to reach position n at which k is placed we have to
> > trverse n positions from head or not...??
> > and i didnt understand the use of ever increasing...please elaborate on it..
>
> > On Tue, Jul 19, 2011 at 2:08 AM, Dumanshu <duman...@gmail.com> wrote:
> > > @Swathi: plz give the TC of ur algo (exponential plus log)
>
> > > On Jul 18, 10:14 pm, Swathi <chukka.swa...@gmail.com> wrote:
> > >> The solution to this problem will be a combination of exponential
> > >> increase
> > >> and the binary search..
>
> > >> start = 0;
> > >> end = 0;
> > >> index =0;
> > >> middle = 0;
> > >> while (1)
> > >> {
> > >> if(a[start] == data)
> > >> return start;
> > >> if(a[end] == data)
> > >> return end;
> > >> if(data > end)
> > >> {
> > >> start = end;
> > >> end = pow(start,2); // here we are consider exponential for faster
> > >> search.
> > >> // no need to check any boundary conditions as the array is infinite
> > >> continue;
> > >> }
> > >> else
> > >> {
> > >> // do binary search between start index and end index because data is
> > >> inbetween a[start] and a[end]
> > >> }
>
> > >> }
>
> > >> Hope this helps...
>
> > >> On Mon, Jul 18, 2011 at 10:32 PM, Gaurav Popli
> > >> <gpgaurav.n...@gmail.com>wrote:
>
> > >> > 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
>
> > >> > > --
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> > >> > > SAGAR PAREEK
> > >> > > COMPUTER SCIENCE AND ENGINEERING
> > >> > > NIT ALLAHABAD
>
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