Delete Nodes From Linked List Present in Array

class Solution {
public:
    ListNode* removeElements(vector<int>& elements, ListNode* root) {
        unordered_set<int> setToRemove(elements.begin(), elements.end());

        while (root != nullptr && setToRemove.count(root->val) > 0) {
            ListNode* toDelete = root;
            root = root->next;
            delete toDelete;
        }

        if (root == nullptr) {
            return nullptr;
        }

        ListNode* ptr = root;
        while (ptr->next != nullptr) {
            if (setToRemove.find(ptr->next->val) != setToRemove.end()) {
                ListNode* toDelete = ptr->next;
                ptr->next = ptr->next->next;
                delete toDelete;
            } else {
                ptr = ptr->next;
            }
        }

        return root;
    }
};

This solution in C++ focuses on removing nodes from a linked list based on a set of values provided in an array. Here's a concise explanation of how the implementation achieves this:

• First, convert the array of elements to be removed into an unordered set. This set aids in quick lookup operations to check if a node's value is in the delete list.
• Begin by checking the head of the linked list. If the head node (root) has a value present in the setToRemove, delete this node and move to the next. Repeat this process until a node not in the set is found or the end of the list is reached.
• After ensuring the new head of the list is not meant to be deleted, traverse the remainder of the list using a pointer. Whenever a node's next node is in the setToRemove, bypass and as a result, delete the node from the list.
• Continue the traversal until the end of the list is reached, ensuring all specified elements are effectively removed.

Remember the importance of proper memory management; after detaching a node from the linked list, delete it to free up memory. This method effectively modifies the linked list in-place, returning the modified list where all nodes matching the values in the array are removed.

class Solution {

    public ListNode filterList(int[] filterNums, ListNode root) {
        // Collect filter numbers into a set for quick access
        Set<Integer> filterSet = new HashSet<>();
        for (int num : filterNums) {
            filterSet.add(num);
        }

        // Remove nodes from the beginning that are in the filter set
        while (root != null && filterSet.contains(root.val)) {
            root = root.next;
        }

        // Return null if the entire list was filtered out
        if (root == null) {
            return null;
        }

        // Delete subsequent nodes containing values in the filter set
        ListNode ptr = root;
        while (ptr.next != null) {
            if (filterSet.contains(ptr.next.val)) {
                // Point to the next next node to effectively remove the current next
                ptr.next = ptr.next.next;
            } else {
                // Otherwise just proceed forward
                ptr = ptr.next;
            }
        }

        return root;
    }
}

This Java solution focuses on removing nodes from a linked list if their values exist within a given array. The approach uses a HashSet to store all elements from the array for quick look-up times, ensuring an efficient filtering of the linked list.

• Initially, populate the HashSet with values from the filterNums array for fast Lookup.
• Start by removing any nodes from the beginning of the linked list that match the values in the HashSet.
• If the entire list is filtered out (i.e., all nodes match and are removed), return a null pointer.
• For remaining elements in the list, traverse through and continue removing any matching nodes by setting pointers to skip over them.

This method ensures that only nodes whose values are not in the filterNums array remain in the linked list, effectively modifying the list in place with a time complexity primarily dependent on the number of elements in the linked list and the operation of searching within the HashSet. By leveraging the HashSet, this search operation for each node is significantly fast, averaging O(1) time.

class Solution:
    def updateList(self, items: List[int], root: Optional[ListNode]) -> Optional[ListNode]:
        # Establish a set from items for quick lookup
        remove_set = set(items)

        # Skip all initial nodes that need to be removed
        while root and root.val in remove_set:
            root = root.next

        # Return None if list becomes empty
        if not root:
            return None

        # Traverse the remaining nodes to remove unwanted values
        current_node = root
        while current_node.next:
            if current_node.next.val in remove_set:
                current_node.next = current_node.next.next
            else:
                current_node = current_node.next

        return root

This summary provides a solution for efficiently removing specific nodes from a singly linked list based on a list of values, using Python 3.

Problem Concept: Given a linked list and a list of integer values (items), the task is to delete any node in the linked list whose value appears in items.

Solution Approach: The solution involves the following steps:

1. Convert the list of integers (items) into a set named remove_set to allow fast lookup operations.

2. Use a while loop to skip and remove all nodes from the beginning of the linked list that are present in remove_set.

3. Check if the linked list is empty after initial removals. If yes, return None as there are no nodes left.

4. Traverse through the linked list starting from the first node that's not in remove_set. For each subsequent node, check if its value is in remove_set.

5. If a node’s value is found in remove_set, bypass this node by adjusting the next pointer of the current node to skip the unwanted node.

6. Continue the traversal until the end of the list is reached.

By completing these steps, all nodes containing values present in the items list will be removed from the linked list, effectively updating it according to specified criteria. This approach ensures an optimal solution using the fast membership testing feature of Python's set data structure during the traversal of the linked list.