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Lesson 37

Reversing Linked List Nodes in K-sized Groups

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Welcome to our next lesson! Last time, we explored how to determine if a linked list is a palindrome, culminating in an optimal O(1) space solution. That exercise gave us valuable practice in combining multiple pointer techniques: finding the middle of a list and reversing a sublist in-place. We'll be putting those pointer manipulation skills to the test today.

This lesson focuses on a challenging and highly-regarded interview problem: reversing the nodes of a linked list in groups of size K. This task requires precise control over pointers to reverse distinct segments of the list and then carefully stitch them back together. Mastering this problem is an excellent indicator of your proficiency with linked lists.

Our primary goal is to develop and understand the optimal iterative solution, which achieves this in O(1) auxiliary space. We will break the problem down into manageable steps, focusing on the logic of identifying, reversing, and reconnecting each group.

1. Understanding the Problem

The task is to take a linked list and an integer k, and reverse the list in k-sized chunks. For example, if the list is 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 and k = 3, the output should be 3 -> 2 -> 1 -> 6 -> 5 -> 4 -> 7. The last group (7) has fewer than k nodes, so it remains unchanged.

Reverse Linked List in groups of Size K - Tutorial

This section from the takeUforward article clearly defines the problem with illustrated examples.

Read the problem definition and examples to ensure you have a solid grasp of the requirements, especially the rule about the final, potentially smaller group.

A straightforward but non-optimal way to approach this would be to extract the node values into an array, reverse the sub-arrays, and then update the values in the linked list. However, this uses O(n) extra space and often violates a common interview constraint: "You may not alter the values in the list's nodes, only nodes themselves may be changed." Therefore, we will focus directly on the optimal, in-place pointer manipulation method.

2. The Iterative In-Place Reversal Strategy

The optimal solution involves traversing the list and, for each k-sized group we find, reversing it and connecting it to the previous and next parts of the list. This sounds complex, so let's break it down.

The main challenges are:

  1. Identifying groups: How do we know if there are at least k nodes left to form a full group?
  2. Reversing a group: How do we reverse just one segment of the list?
  3. Reconnecting: How do we link the end of the previous reversed group to the start of the current one, and the end of the current one to the start of the next?

We can solve this systematically using a few key pointers and a helper function. The logic is greatly simplified by using a dummy node at the beginning of the list. This gives us a fixed predecessor for the very first group, avoiding edge cases for updating the list's head.

The overall algorithm will be:

  1. Create a dummy node that points to the head.
  2. Use a pointer, let's call it groupPrev, initialized to dummy. This pointer will mark the tail of the last processed group.
  3. Loop as long as there are full k-groups remaining:
    a. From groupPrev, find the k-th node of the current group. If it doesn't exist, we're done with reversals.
    b. Identify the start of the next group (groupNext).
    c. Reverse the k nodes starting from groupPrev.next.
    d. Reconnect the pointers: groupPrev should now link to the new head of the reversed group, and the new tail of the reversed group should link to groupNext.
    e. Advance groupPrev to the tail of the group we just reversed, preparing for the next iteration.
  4. Return dummy.next, which is the new head of the fully modified list.

The following image provides a brilliant step-by-step visualization of this entire process for k=3. Pay close attention to how the pointers (temp, kthNode, prevLast, nextNode) are managed to reverse each group and then stitch the list back together.

This diagram illustrates the iterative reversal of a linked list in groups of three. It shows the initial state, finding the k-th node, reversing the group, and then reconnecting it before moving to the next group.

To see this logic animated and explained in detail, the following video is an excellent resource.

L21. Reverse Nodes in K Group Size of LinkedList

This video from the takeUforward channel provides a clear, step-by-step walkthrough of the optimal iterative solution. It covers identifying the k-th node, reversing the sublist, and reconnecting the parts.

Watch from the beginning to understand the overall strategy. Pay special attention to these key segments: Reversing the first group and updating the head. Connecting subsequent groups, where the concept of a "previous node" to link to is introduced. Translating the logic into pseudocode, which solidifies the algorithm's structure.

Now that you have the conceptual and visual understanding, let's examine a clean Java implementation.

Reverse Linked List in groups of Size K - Tutorial

This article provides the full algorithm and a well-structured Java implementation for the optimal approach.

First, read the Algorithm section which outlines the steps, including the use of helper functions. Then, carefully study the Java code provided. Notice how the code is structured: A getKthNode helper function cleanly handles the task of finding the end of a group. A dummy node simplifies the connections. The main while loop orchestrates the process of finding a group, reversing it, and re-linking it. The reversal itself is done inside the loop with a for loop that runs k times.

3. Complexity Analysis and Alternative Approaches

The iterative solution we've developed is the standard for interviews because of its efficiency.

  • Time Complexity: O(N). Although we have nested loops, each node is visited a constant number of times. We traverse once to find the kth node and once to reverse it. So, the total complexity is linear.
  • Space Complexity: O(1). We only use a handful of pointers (dummy, groupPrev, kth, etc.), regardless of the list size.

There is also a recursive way to solve this problem, which can be more elegant to write but has a space complexity drawback.

Reverse Nodes in k-Group | DSA | AlgoMaster.io

The AlgoMaster.io article provides a good summary of the recursive solution.

Read the section on Approach 3. The core idea is to reverse the first k nodes and then recursively call the function on the rest of the list. Understand why this leads to O(n/k) space complexity due to the call stack.

While the recursive solution is a valid way to think about the problem, the iterative O(1) space solution is what you should aim to master for technical interviews.

4. Practice Problem

You are now ready to implement this yourself. This is classified as a 'Hard' problem on LeetCode, so take your time and apply the structure we've discussed. Use a dummy node and helper functions to keep your code clean.

Conclusion

In this lesson, we tackled one of the most involved linked list problems, building directly on the reversal skills from our previous session.

Key Takeaways:

  • Decomposition is Key: Complex linked list problems can be solved by breaking them down into smaller, manageable tasks: finding a sublist, reversing it, and reconnecting it.
  • Dummy Nodes Simplify Logic: Using a dummy node is a powerful pattern that eliminates the need to handle the head of the list as a special case, especially when the head itself is being modified.
  • Pointer Management: The solution hinges on carefully managing multiple pointers to keep track of the boundaries between groups (groupPrev, kth, groupNext).
  • Iterative vs. Recursive: The iterative O(1) space solution is the optimal and expected approach in an interview setting, while the recursive solution offers an alternative but less space-efficient way to frame the logic.

This lesson concludes our deep dive into linked list manipulations. In our next module, we will explore two fundamental linear data structures: stacks and queues. We will start by implementing them from scratch to understand their core Last-In, First-Out (LIFO) and First-In, First-Out (FIFO) principles.