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Intuition\nThe main idea behind this code is to modify a linked list by removing any node whose value appears in a given array nums. following solution efficien
# Intuition\nThe main idea behind this code is to modify a linked list by removing any node whose value appears in a given array nums. following solution efficiently eliminates unwanted nodes from the linked list by leveraging a set for quick lookups and adjusting the list\'s pointers accordingly\n\n# Approach\n# Approach for the code:\n\n1. **Convert `nums` to a Set**:\n   - First, convert the given `nums` array into a **Set** (`numsSet`) to optimize lookup operations. Using a set allows you to check if an element exists in constant time (O(1)). This step ensures that when you traverse the linked list, you can quickly determine whether a node\u2019s value is present in the set.\n\n2. **Use a Dummy Node**:\n   - A **dummy node** is initialized with a default value of `0` and its `next` pointer set to the head of the original list. This dummy node helps to manage edge cases, especially when the head node needs to be removed. It simplifies pointer manipulation and avoids the need for special handling of the first node.\n\n3. **Iterate over the Linked List**:\n   - The iteration begins from the dummy node and proceeds through the list. For each node:\n     - Check if the value of the next node (`curr.next.val`) exists in the `numsSet`.\n     - If it does, remove the node by updating the `next` pointer to skip the current node (`curr.next = curr.next.next`).\n     - If the value is not in `numsSet`, move to the next node (`curr = curr.next`).\n   \n4. **Return the Modified List**:\n   - Once the loop finishes, return the modified list starting from `dummy.next` (skipping the dummy node). This returns the modified version of the list with the desired elements removed.\n\nThis approach efficiently removes nodes by leveraging a set for fast lookups, and uses pointer manipulation to skip over nodes that need to be removed. The overall complexity is O(n + m), where n is the number of elements in `nums` and m is the number of nodes in the linked list.\n\n\n# Complexity\n- Time complexity:\nO(\u2223nums\u2223+\u2223head\u2223)\n\n- Space complexity:\n- O(\u2223nums\u2223)\n\n# Code\n```java []\nclass Solution {\n  public ListNode modifiedList(int[] nums, ListNode head) {\n    ListNode dummy = new ListNode(0, head);\n    Set<Integer> numsSet = Arrays.stream(nums).boxed().collect(Collectors.toSet());\n\n    for (ListNode curr = dummy; curr.next != null;)\n      if (numsSet.contains(curr.next.val))\n        curr.next = curr.next.next;\n      else\n        curr = curr.next;\n\n    return dummy.next;\n  }\n}\n```
['Array', 'Hash Table', 'Linked List', 'Math', 'Java']