Final answer:
The worst-case and average-case running times are generally the same for operations in a singly linked list, except when adding an item to the end if no tail reference is kept; in this specific case, the worst-case running time is O(n) while the average case might be incorrectly assumed to be O(1). The correct option is D.
Step-by-step explanation:
The student's question pertains to the complexity of various operations in a singly linked list. When considering worst-case and average-case running times for operations on a singly linked list, these tend to be the same due to the list's structure and access patterns, except in some scenarios.
- Retrieving the first item of the list (A) is an O(1) operation, meaning it takes constant time both in average and worst-case scenarios since it's direct access to the head of the list.
- Retrieving the last item of the list (B) requires traversing the entire list. The worst-case and average-case time complexity is O(n), where n is the number of elements in the list because you have to visit each node to get to the end.
- Adding an item to the beginning of the list (C) is also an O(1) operation. It's a simple matter of updating the head of the list, making it a constant time operation for both worst-case and average-case.
- Adding an item to the end of the list (D), in the worst case, requires traversing the whole list which is an O(n) operation. However, if you keep a reference to the last item, this could be optimized to O(1). Without this optimization, the average case could mistakenly be assumed as O(1) if the list is frequently empty or very short, but the worst case remains O(n).
In summary, the worst-case running time for adding an item to the end of the list is different from the average-case running time if no optimization is used to keep a reference to the last item.