Some please give the TC for exponential + logarithmic search method
pointed out by Swathi.
On Jul 19, 11:51 am, Ankur Khurana <ankur.kkhur...@gmail.com> wrote:
> @sagar: you might have proposed a solution but it ever occured to you that
> (ptr->next)->next means that you have gone through two diferent nodes. For
> saving one comparison , we are doing lot of un necessary computations which
> , IMHO is not good or feasible . Linked list are not made for searching
> purpose, for that arrays were made.
>
> On Tue, Jul 19, 2011 at 12:13 PM, sagar pareek <sagarpar...@gmail.com>wrote:
>
>
>
>
>
>
>
> > hey
> > check my algo...i mentioned all the possible cases :)
>
> > On Tue, Jul 19, 2011 at 11:31 AM, oppilas .
> > <jatka.oppimi...@gmail.com>wrote:
>
> >> Yes we can avoid integer comparison.
>
> >> But for jumping we will need to check whether the next node is null or
> >> not.
> >> So, depending upon the jump size, the number of comparison in worst case
> >> can be very large if our jump size is increasing exponentially. So, why not
> >> just compare with the next node instead of jumping.
>
> >> On Tue, Jul 19, 2011 at 10:47 AM, sunny agrawal
> >> <sunny816.i...@gmail.com>wrote:
>
> >>> yes Alok u r right that in any case we will traverse k times if k is the
> >>> position if the element that need to searched
> >>> but by jumping we can save the time of avoiding unnecessary comparisions
>
> >>> On Tue, Jul 19, 2011 at 10:44 AM, Alok Prakash
> >>> <alok251...@gmail.com>wrote:
>
> >>>> If i am not wrong, in a linked list to move to any number of position,
> >>>> say n, we have to traverse all intermediate nodes of the linked list.
>
> >>>> So, it does not matter if we are moving pointer by 2,4,8,... positions,
> >>>> we have to scan all intermediate nodes.
>
> >>>> Is it not a simple searching????
>
> >>>> On Tue, Jul 19, 2011 at 9:17 AM, sumit <sumitispar...@gmail.com> wrote:
>
> >>>>> 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.
>
> >>>>> > > >> > >>> --
> >>>>> > > >> > >>> You received this message because you are subscribed to
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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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> >>>>> > > >> - Show quoted text -
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> >>>> --
> >>>> Regards
> >>>> Alok Prakash
>
> >>>> --
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> >>> --
> >>> Sunny Aggrawal
>
> ...
>
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