Nth node from end of linked list

/*
Method 1:find length then do len-n and print (len-n)th node
         T.C O(n) but two pass S.C O(1)
Mthod 2:use recursion T.C O(n), S.C O(n) due to stack of recursion calls
Method 3: use two pointers
    first make two pointer.keep first pointer at head and
    move second pointer to nth node.Now first pointer is
    nth left of second pointer.now move both pointer one
    step at a time until second reaches at end.
    still first pointer will be nth left of second pointer
    and end of linkedlist also.
    T.C O(n) two pass but one pass is not full linked list,S.C O(1)
*/
        fast = head
        slow = head

        # Move fast n steps ahead
        for _ in range(n):
            fast = fast.next

        # If fast is None, that means we need to remove the head
        if not fast:
            return head.next

        # Move both pointers until fast reaches the last node
        while fast.next:
            slow = slow.next
            fast = fast.next

        # Remove the target node
        slow.next = slow.next.next

        return head