in DS
502 views
5 votes
5 votes

You are given a linked list, $L$ of $n$ integers, and another linked list, $P$, of $k$ integers where $k <<< n$. $L$ is sorted in ascending order but $P$ is not necessarily sorted. The operation $\text{print_lots}(L,P)$ will print the elements in $L$ that are in positions specified by $P$. For instance, if $P = 2, 1, 3, 6$, the second, first, third, and sixth elements in $L$ are printed in the same order. If you write the routine function $\text{print_lots}(L,P)$, the best running time of your routine function will be (assume $O(n)$ auxiliary space is available)  ______________ ?

  1. $\Theta(n)$
  2. $\Theta(n \log k)$
  3. $\Theta(n k)$
  4. $\Theta (n k\log k)$
in DS
by
502 views

3 Answers

6 votes
6 votes
Best answer

It looks O(NK) ?

But, it is given that k<<<N and auxiliary space of O(N) is there.

So, sort P array in O(K log K) and copy/store the order of P array in an auxiliary array, in which we have to print the elements of L.

Hence, in one traversal of head pointer throughout the list, this can be done.

Total Time Complexity = O(K Log K)  +  O(K)  + O(N) = O(N) as k<<<<N

selected by
by

4 Comments

yes, I'll do so. As of now, there is no automatic way to do so.
1
1

@ sir

can we use this auxilary space as Array ??

0
0
Why not just use the O(n) space as an auxillary array, copy all linked list elements in the array in O(n) time. Now any element can be accessed directly.
1
1
5 votes
5 votes
We are given with the linked list $L$ which contains $n$ elements. It is also mentioned that we have $\theta (n)$ space.

One thing that we can do is, traverse the whole list $L$, and save it to an array of size $n$, hence using that auxiliary space of $\theta (n)$. This will take $\theta (n)$ time.

Now traverse the list $P$ having $k$ elements, and for each element in $P$ print the element in that index in our array.

Say $P$ is $2,1,6$. Then we start traversing $P$, first is $2$, so we print $2nd$ element of our array, then $1st$ element, and then $6th$ element. And so on. This will take $\theta (k)$.

The whole procedure will take  $\theta (n) + \theta (k)$, ie $ = \theta (n)$. Since, $k<<<n$.

1 comment

I don't know why everybody is discussing so much on the best answer chosen when this is the simplest approach.
0
0
0 votes
0 votes
For each entry in P we have to search Linked List (L).
No of entries in P is 'k' and average search time in Linked List is O(n).

Total time is k*O(n) i.e. O(kn).
Sir why answer is 'n' why not 'nk' ?

4 Comments

awaiting!
0
0
is this method correct? Traverse linked list L and copy the elements in the array and since linked list is sorted in ascending order, array will also be sorted. Then index the array for the positions specified by the linked list p and print them. It will take O(n) time.
0
0
@arjun sir i got the solution of Risabh Gupta but i didn't get the best answer if we sort it then it Is O(klogk) and for copy into auxiliary space it will take O(k) but then how we can print the list elements in one traversal of head pointer ?
0
0
Answer:

Related questions