A meld operation on two instances of a data structure combines them into one…
2025
A meld operation on two instances of a data structure combines them into one single instance of the same data structure. Consider the following data structures:
P: Unsorted doubly linked list with pointers to the head node and tail node of the list.
Q: Min-heap implemented using an array.
R: Binary Search Tree.
Which ONE of the following options gives the worst-case time complexities for meld operation on instances of size 𝑛 of these data structures?
- A.
P: Θ(1), Q: Θ(𝑛), R: Θ(𝑛)
- B.
P: Θ(1), Q: Θ(𝑛 log 𝑛), R: Θ(𝑛)
- C.
P: Θ(𝑛), Q: Θ(𝑛 log 𝑛), R: Θ(𝑛2)
- D.
P: Θ(1), Q: Θ(𝑛), R: Θ(𝑛 log 𝑛)
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Correct answer: A
Final answer: P: Θ(1), Q: Θ(n), R: Θ(n).
P (unsorted doubly linked list with head and tail pointers): Meld by linking the tail of the first list to the head of the second and updating head/tail pointers (and handling empty lists). This is a constant number of pointer updates, so Θ(1).
Q (min-heap implemented using an array): Concatenate the two underlying arrays into one array of size 2n and run build-heap (heapify) on the combined array. Heapify runs in linear time in the array size, so the cost is Θ(2n)=Θ(n). (Note: inserting elements one-by-one would be Θ(n log n), but that is not the optimal meld algorithm.)
R (binary search tree): Perform inorder traversal of each tree to produce two sorted lists (Θ(n)), merge the two sorted lists into one sorted list (Θ(n)), and build a BST from the merged sorted list (Θ(n)). Totals to Θ(n). (A naive approach inserting nodes one-by-one into an unbalanced tree could be Θ(n^2), but the optimal melding procedure is linear.)
Therefore the worst-case time complexities for optimal meld procedures are:
Unsorted doubly linked list with head/tail pointers: Θ(1)
Min-heap in array: Θ(n)
Binary search tree: Θ(n)
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