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