# Solving Leetcode Interviews in Seconds with AI: Remove Nth Node From End of List


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "19" 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, remove the nth node from the end of the list and return its head.   Example 1:   Input: head = [1,2,3,4,5], n = 2 Output: [1,2,3,5]  Example 2:  Input: head = [1], n = 1 Output: []  Example 3:  Input: head = [1,2], n = 1 Output: [1]    Constraints:  The number of nodes in the list is sz. 1 <= sz <= 30 0 <= Node.val <= 100 1 <= n <= sz    Follow up: Could you do this in one pass? 

	# Explanation
	Here's the breakdown of the problem and the solution:

*   **High-Level Approach:**

    *   Use two pointers, `slow` and `fast`, initialized to the head of the list.
    *   Move the `fast` pointer `n` nodes ahead. This creates a gap of `n` nodes between `slow` and `fast`.
    *   Move both `slow` and `fast` pointers simultaneously until `fast` reaches the end of the list. At this point, `slow` will be pointing to the node just *before* the node to be removed. Remove the next Node.

*   **Complexity:**

    *   Runtime: O(N), where N is the number of nodes in the list (one pass).
    *   Storage: O(1) (constant extra space).

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

def removeNthFromEnd(head, n):
    """
    Removes the nth node from the end of a linked list.

    Args:
        head: The head of the linked list.
        n: The position of the node to remove from the end.

    Returns:
        The head of the modified linked list.
    """

    # Create a dummy node to handle the case where the head needs to be removed.
    dummy = ListNode(0)
    dummy.next = head
    slow = dummy
    fast = head

    # Advance 'fast' pointer n nodes ahead
    for _ in range(n):
        if not fast:  # Handle edge case where n > list length. Although prompt specifies 1 <= n <= sz, it's good to include a sanity check
            return head
        fast = fast.next

    # Move both pointers until 'fast' reaches the end
    while fast:
        slow = slow.next
        fast = fast.next

    # Remove the nth node from the end
    slow.next = slow.next.next

    return dummy.next
	```