How would you implement a function to remove the last node from a singly linked list with a tail pointer?

• A. Even with a tail pointer, removing the last node typically requires locating the second-last node to update its next to NULL and updating tail.
• B. Tail pointer allows O(1) removal of last node by updating tail to NULL.
• C. You can just set tail.next = NULL.
• D. You must traverse from tail backwards.

Answer: (A)

Removing the last node in a singly linked list with a tail pointer hinges on not having a direct pointer to the node before the last. In a singly list, each node points forward, and the tail points to the final node whose next is NULL. To remove that last node, you need to detach it by changing the next field of the node that comes just before it to NULL, and then update the tail to that preceding node. Since there’s no backward link, you typically start at the head and walk forward until you find the node whose next is the tail. If the list has only one node, you simply set both head and tail to NULL. This operation is O(n) in the general case because of the traversal, unless you maintain an extra pointer to the previous node or convert to a doubly linked list. So the correct approach is to locate the second-last node, set its next to NULL, and move the tail to that node.