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.
>> >
>> > > --
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>> >
>> > --
>> > Regards
>> > Kumar Raja
>> > M.Tech(SIT)
>> > IIT Kharagpur,
>> > [email protected]
>> > 7797137043.
>> > 09491690115.- Hide quoted text -
>> >
>> > - Show quoted text -
>>
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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.

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