> For the complete documentation index, see [llms.txt](https://private-26.gitbook.io/notes/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://private-26.gitbook.io/notes/coding/hard/25.-reverse-nodes-in-k-group.md).

# 25. Reverse Nodes in k-Group

The goal is to reverse the nodes of a linked list in groups of size `k`. If the remaining number of nodes is less than `k`, they should stay in their original order.

We do this by identifying chunks of size `k`, reversing them, and reconnecting the reversed parts back into the main list.

For each group of size `k`:

* Find the `groupEnd`.
* Temporarily break the connection to the rest of the list.
* Reverse the group.
* Reconnect the reversed group to the list.
* Update `prevGroupTail` and `groupStart` to process the next group.

## Complexity

| Space Complexity | Time Complexity |
| ---------------- | --------------- |
| $$\text{O}(1)$$  | $$\text{O}(N)$$ |

## Code

```java
// Helper function to reverse a linked list starting from 'node'
void reverse(ListNode node) {
    ListNode previous = null;
    while (node != null) {
        ListNode nextNode = node.next;
        node.next = previous;
        previous = node;
        node = nextNode;
    }
}

public ListNode reverseKGroup(ListNode head, int k) {
    // No need to process if k <= 1 or list is empty
    if (k <= 1 || head == null) return head;

    // Dummy node simplifies edge cases (e.g. reversing at head)
    ListNode dummyNode = new ListNode(-1, head);
    ListNode prevGroupTail = dummyNode;
    ListNode groupStart = head;

    while (groupStart != null) {
        ListNode groupEnd = groupStart;
        int count = k;

        // Find the end node of the current group
        while (count > 1 && groupEnd != null) {
            groupEnd = groupEnd.next;
            count--;
        }

        // If the group is smaller than k, no reversal needed
        if (groupEnd == null) break;

        // Detach the group and save the remaining list
        ListNode nextGroupStart = groupEnd.next;
        groupEnd.next = null;

        // Reverse the current group
        reverse(groupStart);

        // Connect the previous group to the new head of this group
        prevGroupTail.next = groupEnd;

        // Connect the tail of the reversed group to the next part
        groupStart.next = nextGroupStart;

        // Update pointers for next iteration
        prevGroupTail = groupStart;
        groupStart = nextGroupStart;
    }

    return dummyNode.next;
}

```


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