How do you find the kth node from the end of a singly linked list in one pass?

• A. Use two pointers; advance the lead pointer k steps, then move both until lead hits null; the second pointer is kth from the end; O(n)
• B. Compute the length, then move from head length - k steps.
• C. Reverse the list and move k steps from head.
• D. Start at head and count to k.

Answer: (A)

The technique rests on using two pointers with a fixed gap. Start both pointers at the head, advance the lead pointer k steps, then move both pointers forward together until the lead pointer reaches the end. At that moment, the lag pointer sits exactly at the kth node from the end. This is one pass through the list and uses O(1) extra space, since only two pointers are tracked. It avoids computing the full length in a separate pass or altering the list by reversing it. If lead becomes null right after the initial k-step advance, the head is the kth node from the end, which is another way this method handles edge cases.