Remove Tail - codepath/compsci_guides GitHub Wiki

Unit 6 Session 1 (Click for link to problem statements)

TIP102 Unit 5 Session 2 Advanced (Click for link to problem statements)

Problem Highlights

• 💡 Difficulty: Medium
• ⏰ Time to complete: 10 mins
• 🛠️ Topics: Linked List, Debugging, Error Handling

1: U-nderstand

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

• Q: What should the output be when the linked list is empty or contains only one element?
  • A: The function should return None as there is no tail to remove, which indicates an empty list.

HAPPY CASE
Input: head = Node(1, Node(2, Node(3, Node(4))))
Output: 1 -> 2 -> 3
Explanation: The tail node with value 4 is successfully removed, leaving the list correctly adjusted.

EDGE CASE
Input: head = Node(1)
Output: None
Explanation: Removing the tail from a single-node list should result in an empty list, returned as `None`.

2: M-atch

Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.

This problem revolves around bug identification and correction in a singly linked list operation, a common issue in data structure manipulation.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Use print statements to trace the code's behavior at critical steps to locate the bug in the tail removal function.

1) Print the current `node`'s value before moving to the next to ensure we are not already at the last node.
2) Add print statements before altering the next pointer to confirm the correct node is modified.
3) Fix the bug by adjusting the loop condition to stop at the second-to-last node!

4: I-mplement

Implement the code to solve the algorithm.

# Debugged remove_tail function with print statements for clarity
def remove_tail(head):
    if head is None:
        print("List is empty. No tail to remove.")
        return None
    if head.next is None:
        print("Only one node in the list. Removing the node.")
        return None

    current = head
    while current.next.next:  # Stop at the second-to-last node
        print(f"Current Node: {current.value} -> Next Node: {current.next.value}")
        current = current.next
    print(f"Removing tail node with value: {current.next.value}")
    current.next = None
    return head

5: R-eview

Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.

• Test with a multi-node list to ensure the last node is removed correctly.
• Test with a single node to confirm the function handles it without errors.

6: E-valuate

Evaluate the performance of your algorithm and state any strong/weak or future potential work.

Assume N represents the number of nodes in the linked list.

• Time Complexity: O(N) because we traverse the linked list up to the second-to-last node.
• Space Complexity: O(1) because the space usage does not increase with the size of the input.