[bhyun-kim.py, 김병현] Week 7 Solutions

Author: sbk1014

remove-nth-node-from-end-of-list/bhyun-kim.py

@@ -0,0 +1,81 @@
+"""
+19. Remove Nth Node From End of List
+https://leetcode.com/problems/remove-nth-node-from-end-of-list/
+"""
+
+
+from typing import Optional
+
+class ListNode:
+    def __init__(self, val=0, next=None):
+        self.val = val
+        self.next = next
+
+"""
+Solution 1: 
+Two Pass Algorithm 1 
+    - First pass: Count the number of nodes and store the values in the list
+    - Second pass: Build the new list without the Nth node from the end
+
+Time complexity: O(N)
+    - Two passes are required
+Space complexity: O(N)
+    - The list stores the values of the nodes
+    - The new nodes are created to build the new list
+"""
+
+class Solution1:
+    def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
+    
+        vals = []
+
+        while head:
+            vals.append(head.val)
+            head = head.next
+
+        dummy_node = ListNode()
+        tail = dummy_node
+        _n = len(vals) - n
+        vals = vals[:_n] + vals[_n+1:]
+
+        for v in vals:
+            tail.next = ListNode(val=v)
+            tail = tail.next
+        
+        return dummy_node.next
+    
+
+
+"""
+Solution 2:
+Reference: 
+    [1] https://leetcode.com/problems/remove-nth-node-from-end-of-list/editorial/
+    [2] https://www.algodale.com/problems/remove-nth-node-from-end-of-list/
+One Pass Algorithm
+    - Use two pointers to maintain a gap of n nodes between them
+    - When the first pointer reaches the end, the second pointer will be at the Nth node from the end
+
+Time complexity: O(N)
+    - Only one pass is required
+Space complexity: O(1)
+    - No extra space is required
+"""
+
+class Solution2:
+    def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
+    
+        dummy = ListNode()
+        dummy.next = head
+        first = dummy
+        second = dummy
+
+        for _ in range(n+1):
+            first = first.next
+
+        while first:
+            first = first.next
+            second = second.next 
+
+        second.next = second.next.next
+
+        return dummy.next