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A list of $n$ strings, each of length $n$, is sorted into lexicographic order using the merge-sort algorithm. The worst case running time of this computation is

  1. $O (n \log n) $
  2. $ O(n^{2} \log n) $
  3. $ O(n^{2} + \log n) $
  4. $ O(n^{2}) $
in Algorithms
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2
Hello @Arjun sir, the answer for the above is O(n^2logn) can you pls explain how to solve this question ?
26
[1] In merge sort of array of length "n" you get a tree of height log(n).

[2] If single element of array is just an integer then at any level of tree, in worst case there will be n comparisons in merge process.

[3] Now if the single element of array is itself an array of size "m" then in order to find larger of 2 elements you need to compare all the m elements.

[4] So at any any level there are n comparisons and each comparison costs "m", so total nmlog(n)
0
thanks alot @sameer2009 for replying.. i got it.
1
In general merge sort contains "n" elements with "log n"levels and in each level "n" data movements so TC is "n logn"

-->here we have "n" strings each of length"n"and each level we have n*n data movementa so "n^2 logn"...
3

What should the recursive equation?

I thought that complexity is T(n)= 2T(n/2) + n2. But this gives a complexity of O(n2) using Master's theorem. This complexity turns out to incorrect according to the answers.

0
yes this is that i want....?
0
O($n^2$) is indeed the correct answer @gmrishikumar

The others have all got it wrong :)
0

Can someone try to obtain the answer using Recurrence relation,I tried but it was giving O(n^2)

11 Answers

0 votes
merge sort take $mlogm$

total no.=n*n

           =n^{2}.

m=no. of elemets

n^{2}logn^{2}$total no.=n*n=n^{2}.

mlogm=n^{2}logn^{2} =2*n^{2}logn$$
0 votes

Hope this helps:)

0 votes
Actually you can think of it as:

$T(n^{2})=2T(\frac{n^{2}}{2})+n^{2}$
 

Work done will be to merge two strings of $\frac{n^{2}}{2}$.

With condition T(n)=1, because strings of length ‘n’ are already sorted.

Now it is pretty straight forward. $O(n^{2}logn)$
Answer:

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