Let SLLdel be a function that deletes a node in a singly-linked list given a pointer to the node and a pointer to the head of the list. Similarly, let DLLdel be another function that deletes a node in a doubly-linked list given a pointer to the node and a pointer to the head of the list. Let $n$ denote the number of nodes in each of the linked lists. Which one of the following choices is TRUE about the worst-case time complexity of SLLdel and DLLdel?

To delete a node from a linked list:

1. Singly-Linked List (SLLdel):

   • To delete a given node, we must update the next pointer of its predecessor node to bypass it.

   • Since each node in a singly-linked list only has a pointer to its successor, we cannot access the predecessor directly.

   • In the worst case (such as when the node to delete is the last node), we must traverse the list starting from the head pointer to locate the predecessor node. This takes $O(n)$ time.

2. Doubly-Linked List (DLLdel):

   • Each node contains direct pointers to both its predecessor (prev) and successor (next).

   • Given a pointer to the target node, we can directly update the adjacent pointers:

     $$\text{node}\to\text{prev}\to\text{next} = \text{node}\to\text{next}$$

     $$\text{node}\to\text{next}\to\text{prev} = \text{node}\to\text{prev}$$

   • This requires only a constant number of pointer operations and does not require traversing from the head. Hence, the worst-case time complexity is $O(1)$.

Therefore, SLLdelis $O(n)$ andDLLdelis $O(1)$, which corresponds to option(D).