Final answer:
Only inserting an item smaller than all existing items at the beginning of the list and finding the smallest element can be done in O(1) time for a singly linked list sorted in ascending order. Operations A and C require traversal of the list and thus cannot be done in O(1) time.
Step-by-step explanation:
Considering a class List that stores values in a sorted ascending order, which is backed by a singly linked list with a head pointer, certain operations may have different time complexities. Here's what we can deduce about the operations and their possible time complexities:
- Insert an item at the correct location in the list (A): To insert an item at the correct location, we generally need to iterate through the list to find the correct position, which takes O(n) time, so this operation cannot be implemented in O(1) time.
- Insert an item smaller than all items currently in the list (B): Since the list is sorted in ascending order and we have direct access to the head of the list, inserting an item smaller than all existing items means we insert at the head, which can indeed be done in O(1) time.
- Delete an item at the correct location in the list (C): Deleting an item requires finding it first, which takes O(n) time, so it cannot be done in O(1) time.
- Find the smallest element (D): Since the list is sorted, the smallest element is at the head, so finding it is an O(1) operation.
Based on this, the operations that can be implemented in O(1) time are inserting an item smaller than all existing items at the beginning of the list (B) and finding the smallest element (D). Therefore, the correct answer is ii. B and D.