Left rotate Linked List by X in groups of Y nodes

Given a singly linked list and two integers X and Y, the task is to left rotate the linked list by X in groups of Y nodes.

Examples:

Input: 10 -> 20 -> 30 -> 40 -> 50 -> 60 -> 70 -> 80 -> 90 -> 100, X = 2, Y = 4
Output: 30 -> 40 -> 10 -> 20 -> 70 -> 80 -> 50 -> 60 -> 90 -> 100
Explanation: First group of nodes is 10->20->30->40. 
After rotating by 2, it becomes 30->40->10->20. 
Second group of nodes is 50->60->70->80.
After rotating by 2, it becomes 70->80->50->60. 
Third group of nodes is 90->100. 
After rotating by 2, it becomes 90->100.

Input: 40 -> 60 -> 70 -> 80 -> 90 -> 100, X = 1, Y = 3
Output: 70 -> 40 -> 60 -> 100 -> 80 -> 90

Approach: The problem can be solved by using reversal algorithm for rotation based on the below observation:

Let A₁ -> B₁ -> A₂ -> B₂ ->.......-> A_nB_n represents the complete list where, A_iB_i represents a group of Y nodes, and A_i and B_i represents the separate parts of this group at the node of rotation i.e. A_i has X nodes and B_i has (Y - X) nodes.
For all Ai and Bi such that 1 ≤ i ≤ N

• Reverse A and B of size X and (Y-X) nodes to get ArBr, where Ar and Br are reverse of A and B respectively,
• Reverse ArBr of size Y to get BA.

Follow the image shown below for a better understanding

Follow the steps mentioned below to solve the problem:

• Traverse the linked list from start:
  • Select the groups of Y nodes:
  • Use reversal approach for rotation in these groups to left rotate by X positions.
  • Move to the next group of Y nodes.
• Return the final linked list.

Below is the implementation of the above approach: 

Given list
10 20 30 40 50 60 70 80 90 100 
Rotated Linked List
30 40 10 20 70 80 50 60 90 100 

Time Complexity: O(N) where N is the length of the linked list
Auxiliary Space: O(1)