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In the worst case, the number of comparisons needed to search a single linked list of length $n$ for a given element is

  1. $\log n$
  2. $\frac{n}{2}$
  3. $\log_2 {n} - 1$
  4. $n$
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6 Answers

25 votes
 
Best answer

A & C are not correct as we can not do binary search in Linked list.

B seems like average case, be we are asked for worst case.

Worst case is we do not find the element in list. We might end up searching entire list & comparing with each element. So, answer -> (D). $n$

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5 Comments

@srestha,how more than n is possible?
0
yes not possible
0
@reena_kandari

more than n is only possible when you apply binary search .

otherwise , by linear method it will take n
0
worst case is not just element is not found, if searched element is the last element then also N comparision are required...isn't it..?
1
SAYING WE CANT DO BINARY SEARCH IS WRONG WE CAN DO BUT IT IS NOT EFFICIENT TO DO BINARY SEARCH IN LINKED LIST IT GIVES ALMOST O(N) TC
5
8 votes
Since, It is not possible to do a Binary search in Linked List, the Only solution is linear search so its O(n).

3 Comments

But is it asked about complexity? It asked about number of comparison only.
0
Binary search is possible on linked list but it is not efficient than linear search so we avoid it and use linear search
0
not impossible but time consuming
0
6 votes

 Binary search algorithm is based on the logic of reducing your input size by half in every step until your search succeeds or input  gets exhausted. Important point here is "the step to reduce input size should take constant time". In case of an array, it's always a simple comparison based on array indexes that takes O(1) time.

But in case of Linked list you don't have indexes to access items. To perform any operation on a list item, you first have to reach it by traversing all items before it. So to divide list by half you first have to reach middle of the list then perform a comparison. Getting to middle of list takes O(n/2)[you have to traverse half of the list items] and comparison takes O(1).
Total = O(n/2) + O(1) = O(n/2)

So the input reduction step does not take constant time. It depends on list size. hence violates the essential requirement of Binary search.
thts why d is the answer only in worst case

4 votes

There are n elements so Log2n is not the case in single linked list because binary search not possible so option A and C is wrong

In worst case there are n comparisons so option D is right 

1 comment

In which case searching in single linked list takes n/2- when we get an element in middle of the list.
0
0 votes
In the worst case scenario,either we will find the element in the last node of otherwise we will not find the node after traversal to whole linked list .so to find a node in the worst case we have to traverse all the linked list till the last node ..that's why no. Of comparisons will also be 'n' .and we don't do binary search in Linked list so log n  base 2 can never be the our answer.
–1 vote
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