LeetCode 25: Reverse Nodes in k-Group — Python Solution

Solve LeetCode 25: Reverse Nodes in k-Group in Python with a group boundary reversal approach. The key is to make the state invariant explicit, so the implementation and complexity follow naturally.

This guide paraphrases the task and does not reproduce LeetCode’s prompt. Use the official page for the complete statement, examples, constraints, and submission runner.

What the problem is testing

Find the kth node from the previous group tail. If it exists, reverse exactly that closed group and reconnect both ends.

Algorithm

1. Find the kth node from the previous group tail. If it exists, reverse exactly that closed group and reconnect both ends.
2. Maintain this invariant: Every completed group is reversed and connected, while nodes after the next boundary remain untouched.
3. Continue until every input item or reachable state has been resolved, then return the accumulated result.

Python solution

LeetCode provides the list, tree, or graph node definition used by the method.

from collections import Counter, defaultdict, deque, OrderedDict
import random

class Solution:
    def reverseKGroup(self, head, k):
        dummy = ListNode(0, head)
        group_prev = dummy
        while True:
            kth = group_prev
            for _ in range(k):
                kth = kth.next
                if not kth:
                    return dummy.next
            group_next = kth.next
            prev, current = group_next, group_prev.next
            while current is not group_next:
                following = current.next
                current.next = prev
                prev, current = current, following
            old_start = group_prev.next
            group_prev.next = kth
            group_prev = old_start

Why this is correct

The proof follows the maintained state: Every completed group is reversed and connected, while nodes after the next boundary remain untouched. Each iteration preserves that claim while permanently resolving at least one position, node, interval, or search state. When the loop or recursion ends, every candidate required by the problem has therefore been included or ruled out, so the returned value is correct.

Complexity

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

Edge cases

A final group shorter than k remains in original order.

Tested reference code

This implementation is included in the site’s downloadable 100-solution Python library. The complete suite compiles every solution and runs a behavioral assertion for every problem before publication.

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