UGC NET CSE | December 2012 | Part 3 | Question: 51

Question

Suppose there are $\log_n$ sorted lists of $n \log_n$ element each. The time complexity of producing a sorted list  of all these elements is (use heap data structure)

• A. $O (n \log \log_n)$
• B. $\theta (n \log_n)$
• C. $\Omega (n \log_n)$
• D. $\Omega (n^{3/2})$

Comments

https://gateoverflow.in/784/gate2005_39

Answer 1

Since total elements are logn * n logn = n(logn)^2 . heapify procedure take O(n (logn)^2) . not in option so Omega (nlogn)

Answer 2

We can merge arrays in O($nk\log k$) time using Min Heap. Following is detailed algorithm.

1. Create an output array of size n*k.
2. Create a min heap of size k and insert 1st element in all the arrays into a the heap
3. Repeat following steps n*k times.
     a) Get minimum element from heap (minimum is always at root) and store it in output array.
     b) Replace heap root with next element from the array from which the element is extracted. If the array doesn’t have any more elements, then replace root with infinite. After replacing the root, heapify the tree.

 

Time complexity of producing a sorted list of all these element=O($nk\log k$)

Here K= $\log n$

n=$n\log n$

Time complexity of producing a sorted list of all these element=

O($nk\log k$)=O$(n\log n\log n\log \log n)$

 

Hence,none of these.

 

Comments

will u  plz elaborate how n= n/logn

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sry it was n Log n.

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as per ugc net published key option A is correct answer