What is the time complexity of appending a node to the end of a singly linked list when a tail pointer is maintained?

• A. O(1)
• B. O(n)
• C. O(log n)
• D. O(n log n)

Answer: (A)

Having a tail pointer in a singly linked list means you can append in constant time because you don’t need to walk the list to find its end. To add a new node at the end, you link the current tail’s next to the new node, move the tail pointer to this new node, and set the new node’s next to null. These are a fixed number of pointer updates and a single allocation, all of which don’t depend on how many nodes are already in the list. Even when the list is empty, you initialize both head and tail to the new node in constant time. That’s why the operation runs in O(1). Without a tail pointer you’d have to traverse from the head to reach the last node, costing O(n). The other growth rates don’t apply here since the end append is a direct, constant-time update with a maintained tail.