Remove Nth Node From End of List
Given the head of a linked list, remove the $n$-th node from the end of the list and return its head. Use the two-pointer technique.
Why Interviewers Ask This
Oracle interviewers use this problem to assess a candidate's ability to optimize space complexity while manipulating linked structures. They specifically evaluate whether you can implement the two-pointer technique efficiently without allocating extra memory for arrays or stacks, demonstrating strong algorithmic thinking and mastery of pointer arithmetic in constrained environments.
How to Answer This Question
1. Clarify edge cases immediately: Ask if n is guaranteed to be valid, if the list can be empty, or if n equals the list length. Oracle values candidates who validate inputs before coding.
2. Propose the Two-Pointer Strategy: Explain that instead of calculating total length first (requiring two passes), you will maintain a gap of n nodes between two pointers to achieve the goal in a single pass.
3. Define Pointer Roles: Assign the 'fast' pointer to move n steps ahead first. Then, move both 'slow' and 'fast' pointers one step at a time until 'fast' reaches the end.
4. Handle Removal Logic: Detail how 'slow' will point to the node just before the target. Use a dummy head node to simplify removal logic, especially when deleting the first node.
5. Refine Code Structure: Write clean code handling the dummy node setup, the initial fast movement loop, the synchronized traversal, and the final pointer reassignment.
Key Points to Cover
- Demonstrating understanding of O(1) space complexity constraints
- Using a dummy node to eliminate special case logic for head removal
- Explaining the maintenance of a fixed gap between two pointers
- Validating input parameters before implementing the algorithm
- Executing a single-pass traversal for optimal efficiency
Sample Answer
To solve the 'Remove Nth Node From End of List' problem efficiently, I would prioritize a single-pass solution using the two-pointer technique, which aligns with Oracle's focus on performance optimization. First, I'd address edge cases by checking if the list is null or if n exceeds the list length. To handle the scenario where we might need to remove the head node without special conditional logic, I'll introduce a dummy node pointing to the actual head. This creates a uniform structure for all deletions.
Next, I initialize two pointers, fast and slow, both starting at the dummy node. I advance the fast pointer n steps forward. At this point, there is exactly an n-node gap between the two pointers. Then, I enter a loop where I move both pointers one step at a time until the fast pointer reaches the last node (its next is null). Because of the maintained gap, the slow pointer will now sit exactly at the node preceding the one we need to remove.
Finally, I perform the deletion by updating slow.next to skip the target node (slow.next = slow.next.next). This approach ensures O(N) time complexity and O(1) space complexity, making it ideal for large datasets often encountered in enterprise systems like those Oracle manages.
Common Mistakes to Avoid
- Failing to use a dummy node, leading to complex conditional checks for head deletion
- Calculating total list length first, resulting in inefficient two-pass solutions
- Incorrectly initializing the fast pointer, causing off-by-one errors in the gap calculation
- Not handling the edge case where n equals the list length (removing the head)
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