> For the complete documentation index, see [llms.txt](https://gl01.gitbook.io/gfg-editorials/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://gl01.gitbook.io/gfg-editorials/2023/07-2023-july-20/21-reverse-a-linked-list-in-groups-of-given-size.md).

# 21. Reverse a Linked List in groups of given size

The problem can be found at the following link: [Question Link](https://practice.geeksforgeeks.org/problems/reverse-a-linked-list-in-groups-of-given-size/1)

## My Approach

To reverse the linked list in groups of K, I used a recursive approach.

* I maintained three pointers, `prev`, `curr`, and `temp`, to keep track of the previous, current, and next nodes respectively.
* I also used a variable `it` to keep track of the remaining elements to reverse in the current group.
* I iterated through the linked list, reversing the pointers, and reducing the `it` count until either the group is fully reversed or we reach the end of the linked list.
* I recursively called the function for the remaining linked list and linked it with current head.

## Time and Auxiliary Space Complexity

* **Time Complexity**: `O(N)`, where `N` is the number of nodes in the linked list.
* **Auxiliary Space Complexity**: `O(N/K)`, where `K` is the size of the group to reverse. The recursion stack uses space for `N/K` function calls.

## Code (C++)

```cpp
class Solution {
public:
    struct node *reverse(struct node *head, int k) {
        if (head == NULL)
            return head;

        struct node *prev = NULL, *curr = head, *temp = head;
        int it = k;

        while (it-- && curr) {
            temp = temp->next;
            curr->next = prev;
            prev = curr;
            curr = temp;
        }

        head->next = reverse(curr, k);

        return prev;
    }
};
```

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