# Solving Leetcode Interviews in Seconds with AI: Rotate List


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "61" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> Given the head of a linked list, rotate the list to the right by k places.   Example 1:   Input: head = [1,2,3,4,5], k = 2 Output: [4,5,1,2,3]  Example 2:   Input: head = [0,1,2], k = 4 Output: [2,0,1]    Constraints:  The number of nodes in the list is in the range [0, 500]. -100 <= Node.val <= 100 0 <= k <= 2 * 109  

	# Explanation
	Here's an efficient approach to solve the linked list rotation problem:

*   **Calculate Length and Effective Rotation:** Determine the length of the linked list. Then, compute the effective rotation `k` by taking `k` modulo the length. This handles cases where `k` is larger than the list's length, preventing unnecessary rotations.
*   **Locate New Head and Tail:** Traverse the list to identify the new head and tail after the rotation. The new tail will be the `(length - k - 1)`th node, and the new head will be the node after the new tail.
*   **Rearrange Pointers:** Break the list at the new tail and connect the old tail to the original head.

*   **Time Complexity:** O(N), where N is the number of nodes in the linked list.
*   **Space Complexity:** O(1) - constant extra space is used.

	
	# Code
	```python
	class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def rotateRight(head, k):
    if not head or not head.next or k == 0:
        return head

    # 1. Calculate the length of the linked list
    length = 1
    tail = head
    while tail.next:
        tail = tail.next
        length += 1

    # 2. Calculate the effective rotation value
    k = k % length

    if k == 0:
        return head

    # 3. Find the new tail and new head
    new_tail_index = length - k - 1
    new_tail = head
    for _ in range(new_tail_index):
        new_tail = new_tail.next
        
    new_head = new_tail.next

    # 4. Perform the rotation
    new_tail.next = None  # Break the list
    tail.next = head       # Connect the old tail to the original head

    return new_head
	```