@Kumar. Use a data structure. The comparison is only on the integer.

Dave

On Aug 2, 8:20 am, kumar raja <[email protected]> wrote:
> But how will u construct the heap with the pair of elements at a time like
> (1,A1) .How to do that??
>
> On 2 August 2011 06:18, kumar raja <[email protected]> wrote:
>
>
>
>
>
> > Thanks Dave ..thanks for ur elaborative explanation
>
> > On 2 August 2011 06:09, Dave <[email protected]> wrote:
>
> >> @Kumar: Form a heap with the first three elements and their sources:
> >> (1,A1), (2,A2), (12,A3). The root of the heap is (1,A1), so output the
> >> 1 and replace it with the next number from list A1: (3,A1). Reheapify,
> >> getting (2,A2), (3,A1), (12,A3). Output 2 and replace it with the next
> >> number from list A2: (4,A2). Reheapify, getting (3,A1), (4,A2),
> >> (12,A3). Output 3. etc etc etc. Eventually, the heap contains (9,A1),
> >> (10,A2), (12,A3). Output 9. Since A1 is exhausted, delete the root.
> >> This can be done by copying the last element in the heap into the root
> >> and decreasing the heap size by 1: (12,A3), (10,A2). Reheapify to
> >> (10,A2), (12,A3). Output 10 and delete (10,A2), leaving the heap
> >> (12,A3) with one node only. Output 12, replace it with 14. Output 14,
> >> replace it with 16. Output 16, replace it with 19. Output 19, replace
> >> it with 20. Output 20. The heap is empty so the process ends. Note
> >> that when only one list remains, it can be copied to the output.
>
> >> Dave
>
> >> On Aug 2, 7:04 am, kumar raja <[email protected]> wrote:
> >> > Could u please explain it with the example i have mentioned ...
>
> >> > On 2 August 2011 04:42, Dave <[email protected]> wrote:
>
> >> > > @Kumar: As you said, insert one number from each list into the heap.
> >> > > Accompany the number with an indication of which list it came from.
> >> > > The minimum number is now at the root of the heap. Output it, replace
> >> > > it with the next number in the same list, and restore the heap
> >> > > condition. When a list is exhausted, delete its entry from the heap
> >> > > and restore the heap condition with the smaller heap. When the entire
> >> > > heap is exhausted, the sorted data is output. Complexity is k log k +
> >> > > (n-k) log k = n log k.
>
> >> > > Dave
>
> >> > > On Aug 2, 5:38 am, kumar raja <[email protected]> wrote:
> >> > > > Let A1[ ] = {1,3,5,7,9}  its size n1
> >> > > > Let A2 []= {2,4,6,8,10} its size n2
> >> > > > Let A3[]= {12,14,16,19,20} its size n3
>
> >> > > >   k=3 sorted  lists are given to us.
> >> > > > n= n1+n2+n3
> >> > > > If i want to merge them  using Heap how to do that ??
> >> > > > I have an idea ,but i am could not able to understand whether it is
> >> > > > right/wrong and what is its time complexity??
>
> >> > > > solution :
>
> >> > > >   I will take the first element in each list and create a heap and
> >> sort
> >> > > it
> >> > > > out.
>
> >> > > > It has complexity of O(k logk) . the remaining elements are inserted
> >> into
> >> > > > Heap one by one and it will take O(log h) (h is the height of heap)
> >> to
> >> > > > insert the new element.
> >> > > > But the height of the heap is keep on increasing as we insert more
> >> and
> >> > > more
> >> > > > elements into it. So i think 'h' cant be a constant ,what should be
> >> the
> >> > > > value of h???
> >> > > > So the complexity comes down to O( (n-k) log h) .because n-k
> >> elements are
> >> > > > left to be inserted and each insertion takes O(log h) time.
>
> >> > > > I want the correct analysis for this problem .
>
> >> > > > --
> >> > > > Regards
> >> > > > Kumar Raja
> >> > > > M.Tech(SIT)
> >> > > > IIT Kharagpur,
> >> > > > [email protected]
> >> > > > 7797137043.
> >> > > > 09491690115.
>
> >> > > --
> >> > > You received this message because you are subscribed to the Google
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> >> > > "Algorithm Geeks" group.
> >> > > To post to this group, send email to [email protected].
> >> > > To unsubscribe from this group, send email to
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> >> > > For more options, visit this group at
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>
> >> > --
> >> > Regards
> >> > Kumar Raja
> >> > M.Tech(SIT)
> >> > IIT Kharagpur,
> >> > [email protected]
> >> > 7797137043.
> >> > 09491690115.- Hide quoted text -
>
> >> > - Show quoted text -
>
> >> --
> >> You received this message because you are subscribed to the Google Groups
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> >> To post to this group, send email to [email protected].
> >> To unsubscribe from this group, send email to
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>
> > --
> > Regards
> > Kumar Raja
> > M.Tech(SIT)
> > IIT Kharagpur,
> > [email protected]
> > 7797137043.
> > 09491690115.
>
> --
> Regards
> Kumar Raja
> M.Tech(SIT)
> IIT Kharagpur,
> [email protected]
> 7797137043.
> 09491690115.- Hide quoted text -
>
> - Show quoted text -

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