Let's explore how to rotate a linked list to the right by k places.

Naive Approach

The brute force method involves repeatedly moving the last element to the front of the list k times. This is straightforward but inefficient.

1. For k iterations:
2. Traverse to the second-to-last node.
3. Move the last node to the beginning.

This approach modifies the list k times, leading to high time complexity.

• Time Complexity: O(n*k), where n is the number of nodes in the linked list.
• Space Complexity: O(1), as it's an in-place algorithm.

This method is highly inefficient for large values of k.

Optimal Approach

A more efficient approach involves these steps:

1. Calculate the length of the list.
2. Handle the case where k is larger than the list's length.
   • Use k % length to get the effective rotation value.
3. Find the new head and tail.
   • Traverse to the (length - k - 1)th node. Its next node will be the new head.
   • The current tail's next pointer should point to the original head.
4. Set the new tail's next pointer to null.

Edge Cases:

• If the list is empty or has only one element, or if k is zero, return the original list.
• If k is a multiple of the list length, no rotation is needed.

Code Example (Java):

public ListNode rotateRight(ListNode head, int k) {
    if (head == null || head.next == null || k == 0) {
        return head;
    }

    ListNode tail = head;
    int length = 1;
    while (tail.next != null) {
        tail = tail.next;
        length++;
    }

    k = k % length;
    if (k == 0) {
        return head;
    }

    ListNode newTail = head;
    for (int i = 1; i < length - k; i++) {
        newTail = newTail.next;
    }

    ListNode newHead = newTail.next;
    newTail.next = null;
    tail.next = head;

    return newHead;
}

• Time Complexity: O(n), where n is the number of nodes.
• Space Complexity: O(1), as it's an in-place solution.

This approach minimizes the number of traversals, leading to better performance.