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- JAVA please Given main() in the ShoppingList class, define an insertAtEnd() method in the ItemNode class that adds an element to the end of a linked list. DO NOT print the dummy head node. Ex. if the input is: 4 Kale Lettuce Carrots Peanuts where 4 is the number of items to be inserted; Kale, Lettuce, Carrots, Peanuts are the names of the items to be added at the end of the list. The output is: Kale Lettuce Carrots Peanuts Code provided in the assignment ItemNode.java:in java please Implement findTheThird method in linked list that searches the bag for a given entry. If found, - removes the first occurrence - leave the second occurrence intact - then replace third occurrence with the string “Found3rd” - remove the rest of the occurrences Return false if no replacement happened. Otherwise, true. public boolean findTheThird (T entry) Note: You may assume that firstNode is a private data in list which references to first node.ctures a... Overview Plans Resources Status and follow-up Participants More v Consider the implementation of orderedLinkedList class, which statement is correct about the following member function: template bool orderedLinkedList:mystery(const Type& x) const bool y = false%; nodeType *current; current = first; while (current != NULL && ly) if (current->info>= x) y = true; else current = current->link; if (y) y= (current->info == x); return y; Your answer: O The function mystery deletes the first item in the list O The function mystery searches for the item x in the list O The function mystery inserts the item x into the correct order in the list O The function mystery deletes the item x from the list O The function mystery inserts the item x into the end of the list 查
- Alert -Don't submit AI generated answer and i need unique response only if I see plagiarism then I'll reduce rating for sure.Given the source code of linked List, answer the below questions(image): A. Fill out the method printList that print all the values of the linkedList: Draw the linked list. public void printList() { } // End of print method B. Write the lines to insert 10 at the end of the linked list. You must draw the final linked List. Notice that you can’t use second or third nodes. Feel free to define a new node. Assume you have only a head node C. Write the lines to delete node 2. You must draw the final linked list. Notice that you can’t use second or third node. Feel free to define a new node. Assume you have only a head nodeAlert -Don't submit AI generated answer and i need unique response only if I see plagiarism then I'll reduce rating for sure.
- Consider a linked list (SLL or DLL your choice) in which, to delete anelement from the list, we simply mark that node for the deletion.Whenever the number of deleted nodes in the list exceeds the number ofnon-deleted nodes, we simply delete all nodes which were marked fordeletion. Implement the following operations for such a linked list.1. addFirst2. addLast3. delFirst4. delLast5. Delete (mark for deletion) an ith node in the list.6. display all non-deleted nodes in the list7. Size : number of non-deleted nodes in the list.8. Recall the last deleted node in the list.Use a Doubly Linked List to implement a Dequea. Define a Deque interface.b. Define a LinkedDeque class (Node class)..c. Define a Test class to test all methods help me make a test class first code package Deque; public class DLinkedList<E> { private Node<E> header = null; private Node<E> trailer = null; private int size = 0; public DLinkedList() { header = new Node<>(null, null,null); trailer = new Node<>(null,header,null); header.setNext(trailer); size = 0; } public int getSize() { return size; } public boolean isEmpty() { return size ==0; } public E getFirst(){ if(isEmpty()) { return null; } return header.getNext().getElement(); } public E getLast(){ if(isEmpty()) { return null; } return trailer.getPrev().getElement(); } public void addFirst(E e) { addBetween(e,header,header.getNext()); } public void addLast(E e) { addBetween(e,trailer.getPrev(), trailer); } public E removeFirst() { if (isEmpty()) { return null; } return…8. Write down the insertBefore method which inserts a new element in the list before the node containing the given element. The method takes as parameters a dummy headed doubly linked circular list, the element existing in the list and new element to be added. public void insertBefore (Node head, Object elem, Object newElement) { //to do OR def insertBefore (head, elem, newElement): pass insertBefore (head, 3, 50) Sample Input Sample Output Ox21 2 223 240 O x21 2 22 50 ² 3 2 4 0
- Add more methods to the doubly linked list class then test them• search(e) // Return one node with 3 values (stuID, stuName, stuScore) which matches agiven key e (studentID).• addAfter(e, stuID, stuName, stuScore) //Add a new node with 3 values (stuID,stuName, stuScore) after the node with the key e (studentID).• removeAt(e) //Remove a node which matches a given key e (studentID)• count() //Return a number of nodes of list.• update(stuID, stuName, stuScore) //Update the values of one node first code below package DlinkedList;public class DLinkedList<A,B,C> { private Node<A,B,C> header; private Node<A,B,C> trailer; private int size; public DLinkedList() { header = new Node<>(null, null, null); trailer = new Node<>(null, null, null); header.setNext(trailer); trailer.setPrev(header); } public int getSize() { return size; } public boolean isEmpty() { return size==0; } public A…"""1.In a directed graph, a strongly connected component is a set of vertices suchthat for any pairs of vertices u and v there exists a path (u-...-v) thatconnects them. A graph is strongly connected if it is a single stronglyconnected component.""" from collections import defaultdict class Graph: """ A directed graph where edges are one-way (a two-way edge can be represented by using two edges). """ def __init__(self,vertex_count): """ Create a new graph with vertex_count vertices. """ self.vertex_count = vertex_count self.graph = defaultdict(list) def add_edge(self,source,target): """ Add an edge going from source to target """ self.graph[source].append(target) def dfs(self): """ Determine if all nodes are reachable from node 0 """ visited = [False] * self.vertex_count self.dfs_util(0,visited) if visited == [True]*self.vertex_count: return True…Please fill in all the code gaps if possible: (java) public class LinkedList { privateLinkedListNode head; **Creates a new instance of LinkedList** public LinkedList () { } public LinkedListNode getHead() public void setHead (LinkedListNode head) **Add item at the front of the linked list** public void insertFront (Object item) **Remove item at the front of linked list and return the object variable** public Object removeFront() }