Delete Duplicates in a Linked List

Unit 6 Session 2 (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: 25 mins
• 🛠️ Topics: Linked Lists, Removing Duplicates

1: U-nderstand

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

• Q: What if all elements are duplicates?
  • A: The list should end up empty except for the temp head.

HAPPY CASE
Input: 1 -> 2 -> 2 -> 3 -> 3
Output: 1
Explanation: Duplicates of 2 and 3 are removed, leaving only 1.

EDGE CASE
Input: 1 -> 1 -> 1
Output: (Empty)
Explanation: All nodes are duplicates and are thus removed.

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 involves removing duplicates from a sorted linked list, which is typically tackled using a pointer to track the last unique node and another to skip duplicates.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Use a temp node to handle edge cases easily, and two pointers to manage duplicates.

1) Initialize a temp node with next pointing to head.
2) Use two pointers: one to keep track of the last node before a series of duplicates (prev) and another to traverse and skip duplicates (current).
3) Adjust links to skip over duplicates.
4) Return the modified list starting from temp's next.

⚠️ Common Mistakes

• Failing to properly reconnect the non-duplicate part of the list after skipping duplicates.

4: I-mplement

Implement the code to solve the algorithm.

def delete_dupes(head):
    # Use a temp node to simplify head operations
    temp = Node(0)
    temp.next = head

    # `prev` is the last node before the current sequence of duplicates or unique values
    prev = temp
    current = head

    while current:
        # Move current to skip over all duplicates
        while current.next and current.value == current.next.value:
            current = current.next

        # If `prev.next` is `current`, no duplicates were found between `prev` and `current`
        if prev.next == current:
            prev = prev.next
        else:
            # Otherwise, skip all duplicates
            prev.next = current.next

        # Move current to the next distinct value
        current = current.next

    return temp.next

5: R-eview

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

• Test the function with a list having multiple duplicates to see if they are removed properly.
• Use an edge case with all identical elements to ensure the list is emptied appropriately.

6: E-valuate

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

• Time Complexity: O(N) where N is the number of nodes in the list. Each node is processed once.
• Space Complexity: O(1) since no extra space is used beyond a few pointers for management.