✅✅✅SIMPLE AND EASY SOLUTION✅✅WITH EXPLANATION USING PYTHON✅✅🔥🔥🔥
PLEASE UPVOTE IF THIS HELPED YOU!!!!\n# Intuition\nTo solve this problem, the goal is to traverse the linked list and remove any nodes whose values are present 
# PLEASE UPVOTE IF THIS HELPED YOU!!!!\n# Intuition\nTo solve this problem, the goal is to traverse the linked list and remove any nodes whose values are present in the list nums. Using a set allows us to efficiently check whether a value exists in nums.\n\n# Approach\n- Convert the list nums into a set for fast O(1) lookups.\n- Create a dummy node to simplify edge cases (like removing the head node).\n- Traverse the linked list using two pointers: prev to track the previous node and curr to point to the current node.\n- If curr.val exists in the set, update prev.next to skip over curr, effectively removing it from the list.\n- Return the updated list starting from dummy.next.\n# Complexity\n- Time complexity: O(n), where n is the number of nodes in the linked list. Each node is visited once.\n- Space complexity: O(m), where m is the size of the set created from the list nums.\n\n# Code\n```python3 []\nclass Solution:\n    def modifiedList(self, nums: List[int], head: Optional[ListNode]) -> Optional[ListNode]:\n        if not head:\n            return None\n        num_set = set(nums)\n        dummy = ListNode(0)\n        dummy.next = head\n        prev, curr = dummy, head\n        while curr:\n            if curr.val in num_set:\n                prev.next = curr.next\n            else:\n                prev = curr\n            curr = curr.next\n        return dummy.next\n\n\n```
['Python3']