Add 'Remove Nth Node From End of List'

Author: nickolashkraus

README.md

@@ -34,6 +34,8 @@ A collection of LeetCode solutions
 
 [Product of Array Except Self](./src/product_of_array_except_self.py)
 
+[Remove Nth Node From End of List](./src/remove_nth_node_from_end_of_list.py)
+
 [Reverse Linked List](./src/reverse_linked_list.py)
 
 [Same Tree](./src/same_tree.py)

src/remove_nth_node_from_end_of_list.py

@@ -0,0 +1,77 @@
+"""
+19. Remove Nth Node From End of List
+
+https://leetcode.com/problems/remove-nth-node-from-end-of-list
+
+NOTES
+  * To remove the nth node from the end of a linked list, we can use a two
+    pointers technique. One pointer (p1) starts at the head, while another
+    pointer (p2) starts n nodes ahead of p1. When p2 reaches the end, p1 will
+    be at the nth node.
+
+    Example (where ○ represents null):
+
+          1   2   3   4   5
+      ○ → ● → ● → ● → ● → ● → ○  n = 2
+      ↑   ↑
+          p1,p2
+      prev
+
+      Start by instantiating a head sentinel node. This allows us to more
+      easily handle the case where n is equal to the length of the list.
+
+      Next, advance p2 by n nodes:
+
+          1   2   3   4   5
+      ○ → ● → ● → ● → ● → ● → ○  n = 2
+      ↑   ↑       ↑
+          p1      p2
+      prev
+
+      Next, advance p1 and p2 together, maintaining prev as the node before p1,
+      until p2 reaches the end of the list:
+
+          1   2   3   4   5
+      ○ → ● → ● → ● → ● → ● → ○  n = 2
+              ↑   ↑           ↑
+                  p1          p2
+              prev
+
+      Finally, remove the nth node (p1) by connecting prev to p1.next:
+
+          1   2   3   5
+      ○ → ● → ● → ● → ● → ○  n = 2
+"""
+
+from src.classes import ListNode
+
+
+class Solution:
+    def removeNthFromEnd(self, head: ListNode | None, n: int) -> ListNode | None:
+        if not head:
+            return None
+
+        # Use a head sentinel node to handle removing the first node.
+        prehead = ListNode(-1)
+        prehead.next = head
+        prev = prehead
+
+        p1, p2 = head, head
+
+        # Advance p2 by n nodes.
+        i = 0
+        while p2 and i < n:
+            p2 = p2.next
+            i += 1
+
+        # Advance both pointers until p2 reaches the end, keeping track of the
+        # node before p1.
+        while p2:
+            prev = p1
+            p1 = p1.next
+            p2 = p2.next
+
+        # Remove p1 by connecting prev to p1.next.
+        prev.next = p1.next
+
+        return prehead.next

tests/test_remove_nth_node_from_end_of_list.py

@@ -0,0 +1,37 @@
+"""
+19. Remove Nth Node From End of List
+
+https://leetcode.com/problems/remove-nth-node-from-end-of-list
+"""
+
+from unittest import TestCase
+
+from src.remove_nth_node_from_end_of_list import Solution
+
+from .utils import create_linked_list_from_list, create_list_from_linked_list
+
+
+class TestSolution(TestCase):
+    def test_1(self):
+        exp = [1, 2, 3, 5]
+        l1 = create_linked_list_from_list([1, 2, 3, 4, 5])
+        ll = Solution().removeNthFromEnd(l1, 2)
+        assert create_list_from_linked_list(ll) == exp
+
+    def test_2(self):
+        exp = []
+        l1 = create_linked_list_from_list([1])
+        ll = Solution().removeNthFromEnd(l1, 1)
+        assert create_list_from_linked_list(ll) == exp
+
+    def test_3(self):
+        exp = [1]
+        l1 = create_linked_list_from_list([1, 2])
+        ll = Solution().removeNthFromEnd(l1, 1)
+        assert create_list_from_linked_list(ll) == exp
+
+    def test_4(self):
+        exp = [2]
+        l1 = create_linked_list_from_list([1, 2])
+        ll = Solution().removeNthFromEnd(l1, 2)
+        assert create_list_from_linked_list(ll) == exp