Rotate List
Given the head of a linked list, rotate the list to the right by $k$ places.
Why Interviewers Ask This
Meta evaluates this problem to assess a candidate's ability to manipulate pointer structures efficiently without allocating extra space. They specifically look for optimization skills, as rotating a list naively often leads to O(N*K) complexity. The interviewer wants to see if you recognize the cyclic nature of the operation and can derive an O(N) solution by identifying the new tail node mathematically rather than simulating each rotation step.
How to Answer This Question
1. Clarify constraints: Immediately ask about edge cases like empty lists, k being larger than list length, or k equal to zero. 2. Analyze the pattern: Explain that rotating right by k is equivalent to moving the last k nodes to the front. Note that effective rotations are k modulo n. 3. Optimize strategy: Propose converting the linked list into a cycle temporarily. Connect the original tail to the head to form a loop. 4. Locate pivot: Calculate the new tail position using (length - k % length). Traverse from the start to find this node. 5. Break cycle: Set the next pointer of the new tail to null and return the new head. This approach ensures O(N) time and O(1) space, aligning with Meta's preference for memory-efficient algorithms.
Key Points to Cover
- Recognizing that k rotations can be simplified using modulo arithmetic
- Identifying the cyclic property to avoid repeated pointer shifting
- Achieving O(N) time complexity instead of O(N*K)
- Maintaining O(1) space complexity without auxiliary data structures
- Handling edge cases like empty lists or k exceeding list length
Sample Answer
To solve the Rotate List problem efficiently, I first need to handle edge cases where the list is empty, has one node, or k is zero, returning the head immediately in those scenarios. Next, I calculate the total length of the list by traversing it once. If k is greater than the length, I take k modulo length because rotating by the full length results in the same list. For example, rotating a 5-node list by 7 is the same as rotating by 2. Once I have the effective k, I realize that a naive simulation would be too slow. Instead, I will treat the list as a circle. I connect the current tail node back to the head, forming a cycle. Now, the problem becomes finding the new tail node, which is located at index (length - k % length) - 1. I traverse from the head to this specific node. Finally, I break the cycle by setting its next pointer to null, making the subsequent node the new head. This entire process runs in O(N) time since we traverse the list at most twice, and uses O(1) extra space, which is optimal for linked list manipulation.
Common Mistakes to Avoid
- Simulating each rotation individually, leading to TLE on large inputs
- Forgetting to handle k values larger than the list length
- Creating a new array or list to store rotated elements, wasting space
- Failing to reconnect the tail to the head before breaking the cycle
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