MANG Problem #7: Reverse Nodes in k-Group (Hard)

1. Problem Statement

Mental Model

Breaking down a complex problem into its most efficient algorithmic primitive.

Given the head of a linked list, reverse the nodes of the list k at a time and return the modified list. k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes, in the end, should remain as it is.

Input: head = [1,2,3,4,5], k = 2
Output: [2,1,4,3,5]

2. The Mental Model: The "Chain of Command"

Think of the linked list as a chain of soldiers. You need to reverse sub-units of size k.

Verification: Before you start reversing a unit, you must confirm there are at least k soldiers in that unit.
Reversal: Once confirmed, you reverse only that segment.
Handoff: The last soldier of your reversed unit (who was the first) must now point to the head of the next reversed unit.

This is the most difficult pointer manipulation problem because it requires managing four different boundaries simultaneously: the previous unit's tail, the current unit's head, the current unit's tail, and the next unit's head.

3. Visual Execution (Pointer Surgery)

graph LR
    subgraph "Step 1: Original"
        1((1)) --> 2((2)) --> 3((3)) --> 4((4))
    end
    
    subgraph "Step 2: Reverse first 2"
        2_((2)) --> 1_((1))
    end
    
    subgraph "Step 3: Connect"
        1_ -.-> 4_((4))
        4_ --> 3_((3))
    end

4. Java Implementation

Recursive Solution (Most Intuitive)

public ListNode reverseKGroup(ListNode head, int k) {
    ListNode curr = head;
    int count = 0;
    
    // 1. Find the (k+1)-th node to see if we have enough nodes to reverse
    while (curr != null && count != k) {
        curr = curr.next;
        count++;
    }
    
    // 2. If we found k nodes, reverse them
    if (count == k) {
        // Reverse the rest of the groups recursively
        curr = reverseKGroup(curr, k); 
        
        // Reverse the current group of k nodes
        // This is standard linked list reversal with a twist
        while (count-- > 0) {
            ListNode next = head.next;
            head.next = curr;
            curr = head;
            head = next;
        }
        head = curr;
    }
    return head;
}

5. Verbal Interview Script (Staff Tier)

Interviewer: "How do you handle the case where the remaining nodes are fewer than K?"

You: "That is the most critical boundary condition. In my implementation, I perform a 'Look Ahead' pass before starting any reversal logic. I attempt to move a pointer k steps forward. If I hit a null before reaching k, I return the current head as-is, which naturally terminates the recursion and preserves the order of the tail segment. This ensures that only full groups are reversed. For the reversal itself, I use a variation of the standard in-place algorithm where the 'next' pointer of the tail node is set to the result of the recursive call for the next group, effectively stitching the reversed segments together in a single pass."

6. Performance Nuances (Staff Level)

Recursion Depth: The recursive solution uses $O(N/K)$ stack space. For a list of 1 million nodes and $k=2$, this would cause a StackOverflowError.
Iterative Optimization ($O(1)$ Space): To be truly "Staff-ready," you should mention the iterative approach using a dummy node and four pointers (prevGroupTail, currGroupHead, currGroupTail, nextGroupHead). It is significantly more complex to code bug-free but provides the optimal $O(1)$ space guarantee required for production systems.

7. Comparison: Standard Reverse vs. K-Group

Feature	Reverse List	Reverse K-Group
Complexity	$O(N)$ Time, $O(1)$ Space	$O(N)$ Time, $O(1)$ Space (iterative)
Logic	Simple swap	Group verification + Stitching
Pointers	3 (prev, curr, next)	5+ (tracking boundaries)

6. Staff-Level Interview Follow-Ups

Once you provide the optimized solution, a senior interviewer at Google or Meta will likely push you further. Here is how to handle the most common follow-ups:

Follow-up 1: "How does this scale to a Distributed System?"

If the input data is too large to fit on a single machine (e.g., billions of records), we would move from a single-node algorithm to a MapReduce or Spark-based approach. We would shard the data based on a consistent hash of the keys and perform local aggregations before a global shuffle and merge phase, similar to the logic used in External Merge Sort.

Follow-up 2: "What are the Concurrency implications?"

In a multi-threaded Java environment, we must ensure that our state (e.g., the DP table or the frequency map) is thread-safe. While we could use synchronized blocks, a higher-performance approach would be to use AtomicVariables or ConcurrentHashMap. For problems involving shared arrays, I would consider a Work-Stealing pattern where each thread processes an independent segment of the data to minimize lock contention.

7. Performance Nuances (The Java Perspective)

Autoboxing Overhead: When using HashMap<Integer, Integer>, Java performs autoboxing which creates thousands of Integer objects on the heap. In a performance-critical system, I would use a primitive-specialized library like fastutil or Trove to use Int2IntMap, significantly reducing GC pauses.
Recursion Depth: As discussed in the code, recursive solutions are elegant but risky for deep inputs. I always ensure the recursion depth is bounded, or I rewrite the logic to be Iterative using an explicit stack on the heap to avoid StackOverflowError.

Key Takeaways

MANG Problem #8: Alien Dictionary (Graph TopoSort)
Problem: Reverse a Linked List II (Range Reversal)
Master the underlying data structure invariants.