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.
>> >
>> > > >> > >>> --
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>> > > >> > >> --
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>> > > >> > >> SAGAR PAREEK
>> > > >> > >> COMPUTER SCIENCE AND ENGINEERING
>> > > >> > >> NIT ALLAHABAD
>> >
>> > > >> > > --
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>> > > >> > > COMPUTER SCIENCE AND ENGINEERING
>> > > >> > > NIT ALLAHABAD
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>
>
> --
> Regards
> Alok Prakash
>
>
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--
Sunny Aggrawal
B-Tech IV year,CSI
Indian Institute Of Technology,Roorkee
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