Chat Topic: Tree - #25 (Reverse Nodes in K-Group)

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Below is a detailed, step-by-step teaching approach for LeetCode 25: Reverse Nodes in K-Group, following your requested structure. Afterward, I will also include a Q&A section for active recall, a generalized Q&A, an analogy for a 5-year-old, a discussion of variations/follow-up solutions, and finally a comparison with other questions you shared. Let’s get started!

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Problem Restatement (LeetCode #25: Reverse Nodes in K-Group)

You’re given the head of a singly linked list, and an integer (k). You need to:

1. Reverse the first (k) nodes of the list,
2. Then reverse the next (k) nodes,
3. Continue this process in chunks of (k) nodes,
4. If you have fewer than (k) nodes left at the end, leave them as they are (do not reverse).

Return the modified list.

Example

• Input: head = [1,2,3,4,5,6], k = 3
• Output: [3,2,1,6,5,4]

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Approach to Explanation

We’ll follow this problem-solving flow:

1. Understanding and Visualization
2. Brute Force Approach
3. Optimization
4. Walk-through
5. Implementation

Then we’ll follow up with:

• Active Recall Q&A (specific to the problem)
• Generalized Q&A (common pitfalls, patterns, and recall questions)
• Summary of the problem’s category, brute force vs. optimized solution, and the key optimization steps
• Q&A for that summary
• Variations / Follow-up Interview Questions + example code changes to handle those variations
• Analogy explanation “like you’re 5 years old”
• Comparison with other questions (including how each question belongs to different categories)

Let’s dive into the details.

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1. Understanding and Visualization

What does the problem ask for?

We have to reverse every group (chunk) of (k) nodes in the linked list. For instance, if (k = 3):

• Original list: [1, 2, 3, 4, 5, 6]
• Split into groups of size 3: [1, 2, 3] and [4, 5, 6]
• Reverse each group: [3, 2, 1] and [6, 5, 4]
• Combine them back: [3, 2, 1, 6, 5, 4]

If the last group has fewer than (k) nodes, we leave it as is (no reversal).

Visualization / Diagram

head
  ↓
  1 -> 2 -> 3 -> 4 -> 5 -> 6 -> NULL
Groups of 3:
  (1 -> 2 -> 3) | (4 -> 5 -> 6)

After reversing each group:
  (3 -> 2 -> 1) | (6 -> 5 -> 4)

Result:
  head -> 3 -> 2 -> 1 -> 6 -> 5 -> 4 -> NULL

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2. Brute Force Approach

A simple (but less optimal) method might be:

1. Traverse the linked list and copy all node values into an array.
2. Reverse the values in chunks of (k) within the array.
3. Convert the array back into the linked list.

Complexities:

• You’d spend (O(n)) time copying values into an array, another (O(n)) time rewriting them into the list, so total (O(n)) time complexity.
• But you also use (O(n)) extra space for storing node values.

This works but uses extra space. The question hints at an in-place method possibly with (O(1)) extra space.

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3. Optimization

Key Insight

We know how to reverse a linked list segment in-place by playing with the next pointers. If we can:

1. Identify each chunk of size (k).
2. Reverse that chunk in-place.
3. Connect it back to the next chunk.

Then we only use (O(1)) extra space. The biggest challenge is to:

• Keep track of where each chunk starts and ends,
• Reverse the chunk,
• Re-link to the next chunk.

General Steps

1. Use a pointer (e.g. groupPrev) to track the node before the chunk you want to reverse.
2. Find the (k)-th node from groupPrev, call it kth.
3. Reverse the sub-list between groupPrev->next and kth.
4. Reconnect the reversed chunk back to the main list.
5. Move groupPrev to the end of the newly reversed chunk, and repeat until no more chunks of size (k) remain.

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4. Walk-Through with Example

Let’s walk through the iterative optimized solution with the example [1,2,3,4,5,6] and (k=3):

1. Create a dummy node that points to the head of the list. dummy -> 1 -> 2 -> 3 -> 4 -> 5 -> 6.

   • Let groupPrev = dummy.

2. Find the (k)-th node from groupPrev:

   • You start from groupPrev, move 3 steps forward. That node is 3 (the (k)-th node).
   • Let kth = 3.
   • Let groupNext = kth.next = 4.

3. Reverse the sub-list from groupPrev->next to kth:

   • Sub-list to reverse is [1 -> 2 -> 3].
   • Reverse it to [3 -> 2 -> 1].
   • After reversing, we have:

     dummy -> 3 -> 2 -> 1
              ↑        |
              └────────┘
     

4. Connect reversed chunk to the rest:

   • groupPrev->next should point to the new head of this chunk (3).
   • The last node in the reversed chunk (1) should connect to groupNext = 4.
   • Now list looks like: dummy -> 3 -> 2 -> 1 -> 4 -> 5 -> 6.

5. Move groupPrev to the last node of the reversed segment, which is 1. Repeat:

   • Next chunk is [4, 5, 6].
   • Reverse [4, 5, 6] to [6, 5, 4].
   • Reconnect. Final list: [3, 2, 1, 6, 5, 4].

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5. Implementation (Iterative Approach)

Below is a commented Java solution demonstrating the iterative approach:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */

class Solution {
    
    public ListNode reverseKGroup(ListNode head, int k) {
        // Create a dummy node to facilitate operations.
        ListNode dummy = new ListNode(0, head);
        
        // groupPrev will mark the node right before the sub-list we're about to reverse.
        ListNode groupPrev = dummy;

        while (true) {
            // Step 1: Find the k-th node from groupPrev.
            ListNode kth = getKthNode(groupPrev, k);
            if (kth == null) {
                // No more groups of size k remain
                break;
            }

            // Step 2: Identify the node after the k-th node
            ListNode groupNext = kth.next;

            // Step 3: Reverse between groupPrev->next and kth inclusive
            ListNode prev = groupNext;
            ListNode curr = groupPrev.next;
            while (curr != groupNext) {
                ListNode temp = curr.next;
                curr.next = prev;
                prev = curr;
                curr = temp;
            }

            // Step 4: Reconnect the reversed sub-list back into the main list
            ListNode tmp = groupPrev.next; // This was the head of the sub-list before reversing
            groupPrev.next = kth;         // kth is now the new head of the reversed sub-list

            // Move groupPrev to the end of this newly reversed segment
            groupPrev = tmp;
        }

        return dummy.next;
    }

    // A helper function to find the k-th node from a given start node.
    private ListNode getKthNode(ListNode start, int k) {
        while (start != null && k > 0) {
            start = start.next;
            k--;
        }
        return start;
    }
}

Explanation of Key Parts

• dummy: Helps handle edge cases (e.g., reversing from the very beginning).
• getKthNode: Moves k times from the given start node; if it returns null, we can’t reverse further.
• Reversal: We adopt the classical linked-list reversal approach (using three pointers): prev, curr, temp.

Time Complexity: (O(n)) — Each node is visited and reversed exactly once or twice, but it’s proportional to (n).
Space Complexity: (O(1)) — We do it in-place without extra data structures.

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Active Recall Q&A (Specific to This Problem)

1. Q: What does “reverse nodes in k-group” mean in simple terms?
   A: It means splitting the list into sub-lists of length (k) and reversing the nodes within each sub-list. If the last chunk doesn’t have (k) nodes, we leave it unchanged.

2. Q: Why do we use a dummy node?
   A: A dummy node simplifies pointer manipulation at the beginning of the list and avoids having to handle special cases for the head.

3. Q: How do we know if there are (k) nodes remaining in the sub-list to reverse?
   A: We move a pointer (k) steps from our groupPrev. If we can’t move (k) steps (we reach null), that means fewer than (k) nodes remain.

4. Q: What happens if (k) = 1 or if (k) is larger than the list size?
   A:

   • If (k=1), reversing groups of size 1 effectively changes nothing.
   • If (k) is larger than the list size, we never find a chunk of size (k), so we leave the list as is.

5. Q: How is the time complexity (O(n))? Aren’t we reversing multiple times?
   A: Even though we reverse sub-lists, each node’s pointer is changed a constant number of times. Overall, no node is processed more than a couple of times, so it’s linear in (n).

6. Q: Where do we see the biggest challenge in the pointer manipulations?
   A: Ensuring we connect the reversed sub-lists back into the main list correctly (fixing groupPrev->next and tail->next). Any mistake here can break the chain or lose nodes.

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More Generalized Q&A (Common Pitfalls / Patterns)

These questions are more about the insight or patterns that might help even if you forget the entire context:

1. Q: In general, how do I know if reversing a sub-list in place is feasible?
   A: If each node has a next pointer (singly-linked), we can reverse in-place by adjusting pointers carefully. We just need to be careful about boundaries (start and end of the sub-list).

2. Q: What’s a common pitfall in reversing linked lists?
   A: Losing the rest of the list by breaking pointers incorrectly. One must keep track of temp (the next node) before re-linking curr.next to the prev.

3. Q: How do I handle partial groups in chunk-based reversing problems?
   A: Usually, you detect if the group length is shorter than (k). If so, you leave it unmodified (no reversal).

4. Q: Why do we often keep a prev pointer in iterative linked list manipulations?
   A: So we can properly link newly reversed portions to the rest of the list. If we only had a pointer to the segment’s head, it’s easy to lose track of the predecessor or the tail.

5. Q: How can we debug pointer errors in reversing sub-lists?
   A: The best approach is to draw the pointers step-by-step (like small diagrams). Also, checking each pointer’s final position after a single iteration can reveal if something is incorrectly linked.

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Summary of the Problem’s Category and Techniques

• Category: Although you mentioned “tree,” this particular question (#25) is classically a Linked List manipulation problem (not actually a Tree problem). Perhaps you labeled it under “tree” by mistake, or maybe you’re grouping certain “data structure” problems under the same umbrella. Nonetheless, conceptually, it’s about reversing parts of a linked list.
• Brute Force: Copy the list’s values to an array, reverse in chunks, copy back. Time: (O(n)); Space: (O(n)).
• Optimized: In-place reversal of each (k)-group using pointer manipulation. Time: (O(n)); Space: (O(1)).
• Key to Optimization: Realizing we can chunk the list by finding the (k)-th node from a “group previous” pointer, reverse that chunk, then link it back into the list, all using constant extra space.

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Q&A for the Above Summary

1. Q: What is the data structure used here?
   A: A singly linked list (plus an optional dummy node to manage the head easily).

2. Q: How do we get from brute force to the optimized approach?
   A: By leveraging in-place reversal of a list segment (the classic “reverse a linked list” approach) repeatedly on chunks of size (k), and carefully reconnecting the sub-lists.

3. Q: Which pointer technique is crucial?
   A: Using a dummy node, a pointer to the chunk’s head, a pointer to the chunk’s tail, and a pointer to the next chunk, ensuring we fix the pointers after reversing.

4. Q: Why do we not need extra data structures?
   A: Because we can rearrange the next pointers themselves to handle the reversal in place.

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Variations or Minor Tweaks

Interviewers often ask variations:

1. “Reverse Nodes in K-Group but keep some nodes sorted or skip reversing certain groups.”

   • You’d add a condition to skip reversing if certain criteria are not met. E.g., if the sum of the group < some threshold, skip reversing.
   • The main pointer manipulation logic remains the same, except you conditionally skip reversing that chunk.

2. “Reverse alternate K groups.”

   • Reverse the first group of size (k), skip the next group of size (k) (just move the pointer), then reverse the next group, etc.
   • You’d modify the loop to only do the reversal step every other iteration.

Example Variation Code (Reverse Alternate K Groups)

Below is a sketch in Java. Notice the difference: we add a toggle to skip every other group:

class Solution {
    public ListNode reverseAlternateKGroup(ListNode head, int k) {
        ListNode dummy = new ListNode(0, head);
        ListNode groupPrev = dummy;
        boolean reverseFlag = true;

        while (true) {
            // Find kth node
            ListNode kth = getKthNode(groupPrev, k);
            if (kth == null) break; // no more groups

            ListNode groupNext = kth.next;

            if (reverseFlag) {
                // Reverse the group
                ListNode prev = groupNext;
                ListNode curr = groupPrev.next;
                while (curr != groupNext) {
                    ListNode temp = curr.next;
                    curr.next = prev;
                    prev = curr;
                    curr = temp;
                }
                ListNode tmp = groupPrev.next;
                groupPrev.next = kth;
                groupPrev = tmp;
            } else {
                // Skip reversing: just move groupPrev directly to the kth node
                groupPrev = kth;
            }

            reverseFlag = !reverseFlag; // flip the boolean
        }

        return dummy.next;
    }
    
    private ListNode getKthNode(ListNode start, int k) {
        while (start != null && k > 0) {
            start = start.next;
            k--;
        }
        return start;
    }
}

Detailed Explanation:

• We keep a reverseFlag that determines whether to reverse the current group.
• If reverseFlag is false, we skip the entire reversal process, just move groupPrev to kth.
• This is a common interview “follow-up” variation.

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Analogy for a 5-Year-Old

Imagine you have a line of toy cars on the floor. You group them in sets of (k). For each group of (k) cars, you flip their order around like turning them in reverse. When you run out of enough cars for a group (fewer than (k)), you don’t flip them—they just stay as they are. At the end, your cars are in a funny order where every group of (k) got reversed, but any leftover group didn’t get touched.

This analogy helps intuitively: you’re reversing the order of each group of (k), then connecting them all in a line again.

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Comparison with Other Linked List / DS Problems

You also shared several other problems from different categories:

• #23 Merge K Sorted Lists (Linked List / Heap / Divide & Conquer)
• #146 LRU Cache (Data Structure / Design with HashMap + Doubly Linked List)
• #287 Find the Duplicate Number (Array / Floyd’s Cycle Detection)
• #2 Add Two Numbers (Linked List addition)
• #138 Copy List With Random Pointer (Linked List + HashMap or in-place trick)
• #19 Remove Nth Node From End of List (Linked List manipulation with two pointers)

Compare and Contrast

1. Data Structures:

   • #25, #23, #2, #138, #19: All revolve around Linked List manipulation.
   • #146 LRU Cache: Uses HashMap + Doubly Linked List for O(1) get/put.
   • #287 Find Duplicate: Typically an Array + Cycle Detection or other techniques.

2. Key Insights:

   • Linked List Reversal (#25) vs. Linked List Merge (#23) vs. Removals (#19) vs. Copying (#138). They all revolve around carefully handling pointers.
   • LRU Cache (#146) is more about data structure design (queues, maps, lists).
   • Find Duplicate (#287) can leverage cycle detection in arrays, or negative marking, or a hash set.
   • Add Two Numbers (#2) is about digit-by-digit addition with carry, using pointers.

3. Optimal Approaches:

   • Linked List manipulations often rely on pointer tricks (dummy heads, fast/slow pointers, or partial reversing).
   • LRU specifically needs a doubly linked list and hash map combo.
   • Find Duplicate can use Floyd’s cycle detection for O(1) extra space.

4. Main Pitfalls:

   • Off-by-one errors and pointer mislinks in linked list operations.
   • Not properly resetting or updating your references.
   • Underestimating corner cases: empty lists, single-element lists, or large (k).

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Short Summary of the Key Insight (All Problems)

• Reversing in groups (#25): Identify chunk boundaries, reverse sub-lists in place.
• Merging Sorted Lists (#23): Repeatedly merge two sorted lists or use a min-heap.
• LRU Cache (#146): Use a custom DS with HashMap + Doubly Linked List for O(1) get/put.
• Find Duplicate Number (#287): Tricks like Floyd’s cycle detection or negative marking.
• Add Two Numbers (#2): Simulate addition digit-by-digit, track carry, watch for leftover carry.
• Copy List with Random Pointer (#138): Use a HashMap or an in-place interleaving technique.
• Remove Nth Node (#19): Either 2-pass length calculation or a 1-pass two-pointer technique.

Each problem teaches a characteristic pointer/DS manipulation or an underlying pattern you can remember for future interviews.

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Q&A for the Above Summaries

1. Q: Which data structure is at the heart of #19, #23, #25, #2, #138?
   A: They all extensively use or manipulate Linked Lists.

2. Q: Which data structure emerges for #146 (LRU)?
   A: A combination of Doubly Linked List + HashMap is standard for an LRU Cache.

3. Q: How does #287 differ from these?
   A: #287 is about Arrays and cycle detection (Floyd’s Tortoise and Hare or negative marking) rather than list manipulation.

4. Q: Is the skill needed for all these questions the same?
   A: They all rely on pointer manipulation and data structure fundamentals, but the specifics (like negative marking vs. next-pointer manipulation) differ. However, the mindset of carefully tracking references/pointers is universal.

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Final Takeaways

1. Problem-Solving: Always break problems down: Understand, Visualize, find a Brute Force, see the pattern, optimize with known DS/Algorithms.
2. Pseudocode & Diagrams: Greatly reduce mistakes in pointer manipulation.
3. Linked List Tools: Dummy heads, two-pointer approach, dedicated helper functions for reversing or merging, etc.
4. Long-Term Memory: Reinforce by doing small, targeted Q&A and re-implementing from scratch occasionally.

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I hope this extensive breakdown helps you develop a deep, long-lasting understanding of Reverse Nodes in K-Group (LeetCode #25), and also gives you context on how to compare with other data-structure problems!

You can revisit these summaries and Q&A sets for active recall whenever you want to refresh your memory on linked list manipulations or other problem-solving patterns.

Good luck, and remember: practice, visualize pointers, and do small recalls often!