How can you compute the length of a singly linked list in a single pass?

• A. Initialize count to 0 and traverse from head to null, incrementing count; return count; O(n)
• B. Store a length field in the head node and return it.
• C. Repeatedly remove nodes while counting until empty.
• D. Use binary search on the nodes.

Answer: (A)

To know the length in one pass, you walk from the head through each next pointer until you reach the end (null), incrementing a counter each time you visit a node. This exactly uses a single traversal, visiting every node once, so it measures the total number of nodes directly. The time is proportional to the number of nodes, O(n), and you only need a couple of variables (the current node and the counter), giving O(1) extra space.

Other ideas aren’t as suitable in a pure single-pass count. Storing a length field in the head can give immediate access to the length, but it requires updating that field with every insertion or deletion, which isn’t a fresh single pass to compute length. Removing nodes while counting would destroy the list structure, which isn’t acceptable if you just want the length. Binary search isn’t applicable here because a singly linked list lacks efficient random access; you can’t skip ahead without traversing. So the straightforward single-pass approach—traverse once from head to null, counting nodes—is the correct method.