Which operation on a singly linked list is guaranteed to be O(1) only when a tail pointer is maintained?

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Multiple Choice

Which operation on a singly linked list is guaranteed to be O(1) only when a tail pointer is maintained?

Explanation:
Maintaining a tail pointer changes how you access the end of a list. In a singly linked list, each node points to the next one, so to append a new node at the tail without a tail pointer you must start at the head and walk through every node until you reach the last one. That traversal takes linear time, O(n). But if you keep a pointer directly to the last node, you can link the new node after the current tail and then update the tail pointer, all in constant time, O(1). Other operations don’t depend on the tail pointer in the same way: inserting at the head is O(1) because you just create the new node and re-point the head. Deleting the head is also O(1) since you move the head pointer to the next node. Traversing to the end to compute length is O(n) because you must visit every node, unless you maintain an additional length counter. So, inserting at the tail is the operation that becomes guaranteed O(1) specifically with a tail pointer.

Maintaining a tail pointer changes how you access the end of a list. In a singly linked list, each node points to the next one, so to append a new node at the tail without a tail pointer you must start at the head and walk through every node until you reach the last one. That traversal takes linear time, O(n). But if you keep a pointer directly to the last node, you can link the new node after the current tail and then update the tail pointer, all in constant time, O(1).

Other operations don’t depend on the tail pointer in the same way: inserting at the head is O(1) because you just create the new node and re-point the head. Deleting the head is also O(1) since you move the head pointer to the next node. Traversing to the end to compute length is O(n) because you must visit every node, unless you maintain an additional length counter.

So, inserting at the tail is the operation that becomes guaranteed O(1) specifically with a tail pointer.

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