LeetCode 92: Reverse Linked List II — Python Solution

LeetCode 92: Reverse Linked List II is a Medium linked list problem. This Python walkthrough develops a head insertion reversal solution, ties every code decision to a concrete invariant, and includes the regression check used before publication.

This independent guide paraphrases the task rather than reproducing LeetCode’s prompt. Use the official page for the exact statement, examples, constraints, and submission runner.

Recognizing the pattern

Linked-list solutions depend on preserving reachability while rewiring a constant number of pointers. Dummy nodes remove many head-only special cases.

For this problem specifically, reach the node before the target segment, then repeatedly remove the next segment node and insert it at the segment front. The invariant worth writing beside the code is: After each insertion, the processed portion of the target segment is reversed in place.

Step-by-step algorithm

1. Identify the input state consumed by reverseBetween(head, left, right) and initialize the data required by the head insertion reversal pattern.
2. Reach the node before the target segment, then repeatedly remove the next segment node and insert it at the segment front.
3. After each update, verify the page’s central invariant: After each insertion, the processed portion of the target segment is reversed in place.
4. Finish only after the boundary behavior is covered: Reversing from the head is handled by the dummy node; left may equal right.

Python solution

LeetCode supplies the list, tree, or graph node class referenced by this method. The downloadable test suite includes compatible local node definitions so the implementation can also run outside the judge.

class Solution:
    def reverseBetween(self, head, left, right):
        dummy = ListNode(0, head)
        before = dummy
        for _ in range(left - 1):
            before = before.next
        tail = before.next
        for _ in range(right - left):
            moved = tail.next
            tail.next = moved.next
            moved.next = before.next
            before.next = moved
        return dummy.next

Reading the implementation

The main entry point is reverseBetween(head, left, right). The named working state includes dummy, before, tail, moved; those variables make the head insertion reversal state visible instead of hiding it in incidental control flow.

The implementation uses 2 loops with separate responsibilities, so preprocessing and the main traversal can each stay linear in their own input. In concrete terms, reach the node before the target segment, then repeatedly remove the next segment node and insert it at the segment front.

Correctness argument

Initialization. The data structure starts with exactly the information known before any input element is processed.

Preservation. Reach the node before the target segment, then repeatedly remove the next segment node and insert it at the segment front. Each update records the current item without invalidating earlier decisions; consequently, after each insertion, the processed portion of the target segment is reversed in place.

Termination. The traversal consumes a finite input or finite state space. At the end, the invariant covers the complete input, which is precisely the condition required for the returned result.

Complexity and trade-offs

O(n) time and O(1) extra space. The auxiliary-space figure excludes the returned output unless the output itself is the structure being built.

Copying values into an array makes indexing easy but abandons the intended pointer-space constraint and node identity. That comparison is useful in an interview because it explains why the final implementation is preferable, not merely that it passes.

Regression check

The published implementation belongs to a 100-problem suite that is compiled and exercised behaviorally before deployment.

One reference assertion from that suite is shown below. It targets the normal path while the edge conditions in the next section cover the failure-prone boundaries.

assert listed(Solution().reverseBetween(linked([1,2,3,4,5]),2,4)) == [1,4,3,2,5]

Common mistakes and edge cases

• Problem-specific boundary: Reversing from the head is handled by the dummy node; left may equal right.
• Pattern-level pitfall: Save the next node before changing a link, terminate reused tails, and check whether the original head can be removed.
• Invariant check: after every update, confirm that after each insertion, the processed portion of the target segment is reversed in place.

Interview review checklist

• Explain why head insertion reversal matches the structure of this input.
• State the invariant in one sentence before tracing code: After each insertion, the processed portion of the target segment is reversed in place.
• Derive O(n) time and O(1) extra space from how many times each element or state is visited.
• Test the boundary explicitly: Reversing from the head is handled by the dummy node; left may equal right.

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