Unit 5 Session 2 (Click for link to problem statements)

Understand what the interviewer is asking for by using test cases and questions about the problem.

• What happens if the linked list has only one node?
  • The function should update the head to None, effectively making the list empty since the tail (which is also the head) is being removed.

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Traverse the linked list to find and remove the last node (tail), updating the second-to-last node's next to None.

1) Check if the list is empty or has only one node by examining `head` and `head.next`. 2) If the list has only one node or is empty, set `head` to `None` to represent an empty list. 3) Otherwise, traverse the list to find the second-to-last node using two pointers: `previous` and `current`. 4) Once the last node is found (`current.next` is `None`), set `previous.next` to `None` to remove the tail. 5) There's no need to return a value as the function modifies the list in place.

⚠️ Common Mistakes

• Not handling cases where the list is empty or has a single node, which requires different handling than removing nodes from longer lists.
• Incorrectly maintaining the previous and current pointers could lead to incorrect or failed removals.

def delete_tail(head): # If the list is empty or has only one node if head is None or head.next is None: return None # After removing the tail, the list is empty # Start with the first node previous = None current = head # Traverse the list until the last node while current.next is not None: previous = current current = current.next # Set the second-to-last node's next to None, removing the last node previous.next = None