### Problem You are given the head of a singly linked list and an integer `k` (`k >= 1`). Reverse the nodes of the list **in-place** in groups of size `k`: - The nodes in each consecutive block of `k` nodes should be reversed as a group. - If the number of nodes at the end of the list is **less than** `k`, leave those nodes in their original order. - Return the new head of the modified list. You may assume the list nodes are defined in the usual way, for example: - In C++: - `struct ListNode { int val; ListNode* next; };` - In Java: - `class ListNode { int val; ListNode next; }` #### Example Input list (values): `1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7`, and `k = 3` Output list should be: `3 -> 2 -> 1 -> 6 -> 5 -> 4 -> 7` - The first 3 nodes `(1, 2, 3)` are reversed to `(3, 2, 1)`. - The next 3 nodes `(4, 5, 6)` are reversed to `(6, 5, 4)`. - The last node `7` is fewer than `k` nodes, so it stays as is. #### Constraints - Number of nodes `n` can be up to around `10^5`. - You must operate in-place using `O(1)` extra space (excluding recursion stack if you choose a recursive solution). - Aim for `O(n)` time. Describe your algorithm and then implement a function that performs this transformation.