To delete the nth node from the end of a singly linked list in one pass, which technique is described?

• A. Use two pointers separated by n nodes
• B. Use a single pass with a counter and track the head
• C. Compute the length first with a preliminary pass
• D. Use a stack to hold the last n nodes

Answer: (A)

The technique relies on keeping a fixed distance between two pointers as you traverse the list, which lets you identify the node to delete in one pass. Start with a dummy node before the head to gracefully handle deleting the first node. Move the second pointer forward by n steps so there is a gap of n between the two pointers. Then advance both pointers together until the second pointer reaches the end of the list. At that moment, the first pointer is right before the node that is nth from the end, so you can remove it by rewiring the next reference (firstPointer.next = firstPointer.next.next). This approach does the work in a single traversal and uses constant extra space, and it also cleanly handles deleting the head node thanks to the dummy node.

Other options either require a preliminary pass to compute the total length, or use extra storage like a stack, or rely on counters and would not achieve the single-pass, constant-space deletion as cleanly.