This problem focuses on removing duplicate values from a sorted linked list, requiring pointer manipulation to ensure only distinct values remain. A direct approach uses two pointers to track duplicates and build the result. Consider the trade-offs between time complexity and pointer management when choosing a solution.
Problem Statement
Given the head of a sorted linked list, your task is to remove all nodes that have duplicate numbers, ensuring that only distinct values remain. The modified list should be returned in sorted order.
For example, if the input list is [1,2,3,3,4,4,5], the output should be [1,2,5], as the duplicates '3' and '4' are removed.
Examples
Example 1
Input: head = [1,2,3,3,4,4,5]
Output: [1,2,5]
Example details omitted.
Example 2
Input: head = [1,1,1,2,3]
Output: [2,3]
Example details omitted.
Constraints
- The number of nodes in the list is in the range [0, 300].
- -100 <= Node.val <= 100
- The list is guaranteed to be sorted in ascending order.
Solution Approach
Pointer Traversal with Two Pointers
Use two pointers to traverse the list. One pointer tracks the current node, and the second identifies duplicate sequences. Remove nodes that are part of duplicate groups and ensure the result maintains sorted order.
Predecessor Node Tracking
Track the predecessor node to link it to the first distinct node when duplicates are found. This allows skipping over all duplicate nodes efficiently.
Edge Case Handling
Handle cases where the list has zero nodes or consists entirely of duplicates. These cases require specific checks to avoid unnecessary operations and ensure correct output.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
The time complexity is O(n), where n is the number of nodes in the linked list, as each node is processed once. The space complexity is O(1), as no additional data structures are used, and the list is modified in-place.
What Interviewers Usually Probe
- Does the candidate correctly apply linked list pointer manipulation?
- How does the candidate handle edge cases, such as empty or duplicate-only lists?
- Is the solution optimal in terms of both time and space complexity?
Common Pitfalls or Variants
Common pitfalls
- Overcomplicating the problem with additional data structures.
- Not correctly linking the predecessor node when skipping duplicates.
- Failing to handle edge cases such as empty lists or all duplicates.
Follow-up variants
- Implementing the solution recursively instead of iteratively.
- Using a hash set to track duplicates (though this adds extra space complexity).
- Modifying the list in-place without using extra pointers.
How GhostInterview Helps
- GhostInterview helps by providing step-by-step guidance on handling pointer manipulation and linked list traversal effectively.
- The platform simulates interview scenarios that help improve the candidate’s performance on linked list problems like this one.
- By analyzing common pitfalls, GhostInterview ensures you’re prepared to avoid common mistakes, making you more confident during real interviews.
Topic Pages
Related GhostInterview Pages
- LeetCode Interview Copilot - Use GhostInterview as a live solver when you want direct help with LeetCode-style coding questions.
- Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
- How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.
FAQ
What is the main pattern to solve Remove Duplicates from Sorted List II?
The main pattern involves linked list pointer manipulation, where you use two pointers to identify and skip duplicate nodes.
How do I efficiently remove duplicates from a sorted linked list?
Use two pointers to traverse the list, keeping track of duplicates and skipping over them, modifying the list in-place to ensure sorted order.
What is the time complexity of the optimal solution?
The time complexity is O(n), where n is the number of nodes, as each node is processed once.
How should edge cases, like an empty list or all duplicates, be handled?
For an empty list, simply return null. For all duplicates, ensure that the solution correctly identifies and removes the duplicate nodes, leaving the list empty or with distinct values only.
Can I use a hash set to solve this problem?
Using a hash set could solve the problem but at the cost of additional space complexity. The optimal solution uses constant space.
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