Leetcode 19

Author: Kondareddy1209

Leetcode/Leetcode_19.py

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+from typing import Optional
+
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+
+
+class Solution:
+    def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
+        """
+        LeetCode 19. Remove Nth Node From End of List
+
+        Approach:
+        1. Reverse the linked list.
+        2. The nth node from the end becomes the nth node from the beginning.
+        3. Remove the nth node.
+        4. Reverse the list again to restore the original order.
+
+        Example:
+            Original:
+            1 -> 2 -> 3 -> 4 -> 5
+
+            Reverse:
+            5 -> 4 -> 3 -> 2 -> 1
+
+            Remove 2nd node:
+            5 -> 3 -> 2 -> 1
+
+            Reverse again:
+            1 -> 2 -> 3 -> 5
+
+        Time Complexity:
+            O(n)
+            - First reverse: O(n)
+            - Remove node: O(n)
+            - Second reverse: O(n)
+
+            Overall: O(n)
+
+        Space Complexity:
+            O(1)
+        """
+
+        # -------------------------------
+        # Step 1: Reverse the linked list
+        # -------------------------------
+        prev = None
+        curr = head
+
+        while curr:
+            nxt = curr.next
+            curr.next = prev
+            prev = curr
+            curr = nxt
+
+        # 'prev' is the new head of the reversed list
+        head = prev
+
+        # -----------------------------------------
+        # Step 2: Remove the nth node from the front
+        # -----------------------------------------
+        prev_node = None
+        curr = head
+        index = 1
+
+        while index < n and curr:
+            prev_node = curr
+            curr = curr.next
+            index += 1
+
+        # If removing the first node of the reversed list
+        if prev_node is None:
+            head = head.next
+        else:
+            prev_node.next = curr.next
+
+        # -------------------------------
+        # Step 3: Reverse the list again
+        # -------------------------------
+        prev = None
+        curr = head
+
+        while curr:
+            nxt = curr.next
+            curr.next = prev
+            prev = curr
+            curr = nxt
+
+        # Return the restored list
+        return prev