PPractick

The problem

A video player buffers frames in a singly linked list and, for a special "rewind in chunks" effect, needs to reverse the frames in consecutive groups of k. If fewer than k frames remain at the end, they stay in their original order. Reversing one group is the standard three-pointer reversal; the real challenge is stitching each reversed group back between the previous group's end and the rest of the list, without losing any node.

Example: 1 -> 2 -> 3 -> 4 -> 5, k = 2 → 2 -> 1 -> 4 -> 3 -> 5.

Function signature

Node* reverseKGroup(Node* head, int k);  // k >= 1; returns the new head

Test cases

Input (list, k)	Output list	Explanation
(1 -> 2 -> 3 -> 4 -> 5, 2)	2 -> 1 -> 4 -> 3 -> 5	The last node is a partial group
(1 -> 2 -> 3 -> 4 -> 5, 3)	3 -> 2 -> 1 -> 4 -> 5
(1 -> 2 -> 3 -> 4 -> 5 -> 6, 3)	3 -> 2 -> 1 -> 6 -> 5 -> 4	Two full groups
(1 -> 2, 1)	1 -> 2	k = 1 changes nothing
(1, 2)	1	Fewer than k nodes

(Confirm: only node links change, not values; a final group shorter than k is left as is; O(1) extra memory is expected.)

How interviewers phrase it

• "Reverse the nodes of a linked list k at a time" (LeetCode 25).
• A GeeksforGeeks variant also reverses the final partial group, so ask which rule applies.

Clarifying questions to ask

1. What happens to a final group shorter than k? (Here it stays as is.)
2. Can I change values, or only links? (Only links.)
3. Is O(1) extra space required? (Rules out recursion if strict.)

Follow-ups to expect

1. "How do you know a full group exists?"
2. "How do you reconnect a reversed group?"
3. "Why a dummy node?"
4. "Recursive version?"
5. "Complexity?"

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