Can we merge two unsorted lists in constant time?

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Discussion Overview

The discussion revolves around the problem of merging two unsorted linked lists while aiming for a constant time complexity. Participants explore the feasibility of this approach, considering the characteristics of singly linked lists and the use of pointers to the head and tail of the lists.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions whether it is possible to merge two unsorted lists in constant time, suggesting that traversing the first list would lead to a complexity of $O(n)$.
  • Another participant proposes maintaining a pointer to the end of the first list to facilitate the merge operation.
  • There is a discussion about the necessity of keeping track of both head and tail pointers for efficient list operations.
  • Some participants clarify that in a singly linked list, each element points only to the next element, not the previous one.
  • Concerns are raised about whether the tail pointer should initially point to NULL and the implications of that for merging lists.
  • Participants discuss the need to define head and tail pointers before implementing the merging algorithm and the expected structure of these pointers in code.
  • There is a suggestion that if the tail pointer is not maintained, it would take $O(n)$ time to find the last element when needed.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of maintaining tail pointers and the implications of their positions on the complexity of the merge operation. The discussion remains unresolved regarding the best approach to achieve constant time complexity in merging the lists.

Contextual Notes

Participants note that the lists in question are singly linked, which affects how pointers can be utilized during the merge process. There is also uncertainty about the initial state of the tail pointer and its relevance to the merging algorithm.

  • #31
I like Serena said:
That's it. (Mmm)

Great! Thank you! (Party)
 
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  • #32
Code:
L1_tail.next=L2.head;
L1_tail=L2_tail;

If we would have an array $A$ and we would use a sentinel node, would it be like that? (Thinking)

Code:
A[i].sentinel=A[j].list;
A[i].sentinel=A[j].sentinel
 

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