What is the efficient way to delete the tail node of a doubly linked list when a tail pointer is maintained?

• A. Tail = tail.prev; tail.next = null
• B. Tail = head; head.next = null
• C. Tail.next = null; head = head.next
• D. Create a new tail node and delete old tail

Answer: (A)

When you maintain a tail pointer, removing the last element should be done in constant time by jumping to the node before the tail and detaching the old tail. The usual steps are: set the tail to tail.prev, then set the new tail’s next to null. This uses the double links to reach the predecessor directly and properly terminate the list from the end, without traversing from the head.

A key edge case is when there was only one node. In that situation, tail.prev is null, so after updating the tail you must also set the head (and tail) to null to reflect an empty list.

This approach is more efficient and correct for a doubly linked list with a tail pointer than options that reset the head, manipulate tail.next without moving tail, or create a new node. It cleanly reuses existing pointers to perform the deletion in constant time.