How would you solve the 'Remove Nth Node From End of List' problem: Remove the n-th node from the tail of the list and return its head?

Overview

This problem is a staple in technical interviews, particularly at FAANG companies, designed to assess a candidate's grasp of linked list manipulation, pointer arithmetic, and edge case handling. It requires not just a correct solution but an optimal one in terms of time and space complexity, reflecting the expectation for efficient and robust code in production environments.

Interview Question:

How would you solve the 'Remove Nth Node From End of List' problem: Remove the n-th node from the tail of the list and return its head?

Why Interviewers Ask This:

Interviewers pose this question to evaluate a candidate's foundational understanding of data structures, specifically linked lists, and their ability to design an efficient algorithm. It tests their proficiency with pointer manipulation, a critical skill for low-level system design and performance optimization. Beyond correctness, the question assesses a candidate's capacity for algorithmic optimality, expecting a solution that achieves O(N) time complexity and O(1) space complexity. This demonstrates a strong grasp of resource management, a key aspect of building scalable and maintainable software.

Furthermore, it reveals a candidate's attention to edge cases, such as an empty list, a list with a single node, or removing the head of the list. The ability to systematically identify and handle these scenarios without introducing bugs is paramount for writing production-grade code. It also evaluates their problem-solving methodology, from initial conceptualization to refining an optimal solution, and their capacity to communicate complex ideas clearly. Writing clean, readable, and robust code that accounts for various inputs is a hallmark of a strong software engineer, reflecting their ownership and commitment to code quality.

Expert Answer:

First, the most efficient way to solve this problem is using a two-pointer approach in a single pass. A naive approach might involve two passes: one to find the length of the list, and a second to find and remove the target node. While functionally correct, this is less optimal in terms of time complexity, requiring two full traversals. The single-pass method, on the other hand, leverages two pointers, often called 'fast' and 'slow', to maintain a fixed distance of 'N' nodes between them, achieving O(N) time complexity with a single traversal.

Next, to simplify edge cases, especially when the node to be removed is the original head of the list, it is highly recommended to introduce a dummy node. This dummy node acts as a sentinel, pointing to the original head of the list. By doing so, the node to be removed will never be the actual head from the perspective of our 'slow' pointer, making the deletion logic consistent regardless of the target node's position. We initialize both 'fast' and 'slow' pointers to this dummy node. This design choice significantly improves code readability and reduces the likelihood of bugs related to special head handling, contributing to better maintainability.

Then, we advance the 'fast' pointer 'N + 1' steps ahead. The '+1' is crucial because it ensures that when 'fast' eventually reaches the end of the list, 'slow' will be positioned one node before the node to be removed. For example, if we need to remove the 1st node from the end (N=1), 'fast' moves 2 steps. If N=2, 'fast' moves 3 steps. This creates the necessary gap. After establishing this fixed distance, we simultaneously advance both 'fast' and 'slow' pointers one step at a time.

Finally, we continue advancing both pointers until the 'fast' pointer reaches the end of the list (i.e., 'fast' becomes 'null'). At this precise moment, 'slow' will be pointing to the node immediately preceding the Nth node from the end. We then perform the deletion by updating 'slow.next = slow.next.next'. This effectively bypasses the target node, removing it from the list. The head of the modified list will be 'dummy.next'. This approach not only guarantees O(N) time complexity but also achieves O(1) space complexity as we only use a constant number of pointers, making it an optimal and production-ready solution.

Speaking Blueprint:

[The Hook] This classic linked list problem is best solved with a single-pass, two-pointer approach, which is crucial for optimal performance in a production setting. My strategy focuses on O(N) time and O(1) space complexity while robustly handling all edge cases.

[The Core Execution] I would start by creating a dummy node that points to the head of the list. This simplifies handling cases where the head itself needs to be removed. Then, I initialize two pointers, 'fast' and 'slow', both pointing to this dummy node. I advance the 'fast' pointer 'N + 1' steps ahead. This offset is key: it ensures that 'slow' will be positioned correctly to perform the deletion when 'fast' reaches the end. After establishing this gap, I move both 'fast' and 'slow' forward one step at a time until 'fast' becomes 'null'. At this point, 'slow' will be exactly one node before the Nth node from the end. I then update 'slow.next' to 'slow.next.next', effectively removing the target node. Finally, I return 'dummy.next' as the new head.

[The Punchline] This method guarantees an optimal solution with a single traversal, making it highly efficient. The use of a dummy node significantly improves code clarity and robustness by abstracting away special handling for head removal, demonstrating a clean and production-ready design.

Common Mistakes:

• Off-by-one errors with N: Incorrectly calculating the distance between 'fast' and 'slow' pointers, or the number of steps 'fast' needs to advance initially, leading to removing the wrong node or index out of bounds issues.
• Not handling edge cases: Failing to consider an empty list, a list with only one node, or the scenario where the head node itself needs to be removed. The dummy node pattern is essential here.
• Two-pass solution: While correct, a two-pass approach (first to count length, second to find node) is less optimal than the single-pass two-pointer method, indicating a missed opportunity for efficiency.
• Incorrect pointer re-assignment: After identifying the node to remove, incorrectly updating the 'next' pointer of the preceding node, which can break the list or lead to memory leaks (in languages without automatic garbage collection).
• Returning the wrong head: Forgetting that if the original head was removed, the new head is 'dummy.next', not the original 'head'.
• Modifying the original head directly: Without a dummy node, removing the head requires special logic and careful handling of the 'head' reference, which can be error-prone.