Implement a priority queue using a heapordered binary tree, however instead of using an array, use a triply linked structure. Three connections are needed for each node: two for climbing the tree and one for descending it. Your solution should offer logarithmic running durations for each operation, even if there isn't a set maximum priority-queue size.
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Implement a priority queue using a heapordered binary tree, however instead of using an array, use a triply linked structure. Three connections are needed for each node: two for climbing the tree and one for descending it. Your solution should offer logarithmic running durations for each operation, even if there isn't a set maximum priority-queue size.
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- Write a program for the implementation of a queue using singly linked list. The elements of the queue should be strings of length at most 20 characters. Print the queue and do the following operations on the created queue : (a) Insert a new element. (b) Delete an element from the queue.Implement a priority queue using a heapordered binary tree, but instead of an array, use a triply linked structure. Each node will require three links: two to traverse down the tree and one to traverse up the tree. Even if no maximum priority-queue size is known ahead of time, your implementation should guarantee logarithmic running time per operation.Given a singly linked list, print reverse of it using a recursive function printLinkedList( node *first ) where first is the pointer pointing to the first data node. For example, if the given linked list is 1->2->3->4, then output should be: 4 3 2 1 (note the whitespace in between each data value)
- The implementation of a queue in an array, as given in this chapter, uses the variable count to determine whether the queue is empty or full. You can also use the variable count to return the number of elements in the queue. On the other hand, class linkedQueueType does not use such a variable to keep track of the number of elements in the queue. Redefine the class linkedQueueType by adding the variable count to keep track of the number of elements in the queue. Modify the definitions of the functions addQueue and deleteQueue as necessary. Add the function queueCount to return the number of elements in the queue. Also, write a program to test various operations of the class you defined.Implement a priority list using a heapordered binary tree, but instead of an array, use a triply linked structure. Each node will require three links: two to travel down the tree and one to traverse up the tree. Even if no maximum priority-queue size is known ahead of time, your implementation should ensure logarithmic running time per operation.Create a new Java class in a file named "ListPQ.java" that implements the Queue interface and uses the LinkedList provided by the JCF to implement a priority queue. Note that LinkedList does not automatically order values based on priority, so it will be up to you to make sure that values are removed in priority order. You should try to implement your priority queue as efficiently as possible - basic operations should run in constant or linear time. Your priority queue need only work with integers. Do not modify any of the provided code. //This is ArrayHeap (provided code) import java.util.Arrays; public class ArrayHeap implements Heap { private int[ ] array; private int size; public ArrayHeap() { array = new int[3]; size = 0; } @Override public void add(int value) { // add pt 1 if(size == array.length) { array = Arrays.copyOf(array, size*2); } array[size] = value; // sifting up int child = size; int parent = (child - 1) / 2; while(array[parent] > array[child]) { swap(parent,…
- Implement a Double Array Queue and test it for a very large case (100,000 randomly decided operations of enqueue or dequeue) Your program should compute the number of costly operations and cheap operations Your program should also ask the user about the ratio between enqueue and dequeue operations: The probability of enqueues and dequeues should never be of less than half the other (34% enqueues - 66% dequeues or 66% enqueues - 34% dequeues)Code in Java only In a rooted tree, the lowest common ancestor (or LCA for short) of two vertices u and v is defined as the lowest vertex that is ancestor of both that two vertices. Given a tree of N vertices, you need to answer the question of the form "r u v" which means if the root of the tree is at r then what is LCA of u and v. Input: 4 12 23 14 2 142 242 Output: 1 2IN JAVA This triple-ended queue exercise can be found at https://open.kattis.com/problems/teque. The requirements for this program are compatible with those of the Kattis problem, so a solution should produce correct results when submitted on Kattis. The Triple-Ended QUEue supports four operations: push_back x: inserts the element x at the end of the TEQUE push_front x: inserts the element x at the front of the TEQUE push_middle x: inserts the element x so that the insertion index for x is (k + 1)/2 for a 0-based indexed TEQUE. get i: prints out the ith index element (0-based) of the TEQUE. The TEQUE will be implemented as a linked list. The usual implementation requirements apply along with a screenshot of the results from Kattis. Sample input 10 push_back 1 push_front 2 push_back 3 push_front 4 push_middle 5 push_back 6 get 3 push_middle 7 push_middle 8 get 5 Sample Output 1 1 Subject: Java Programming
- Suppose there are two singly linked lists both of which intersect at some point and become a single linked list. The head or start pointers of both the lists are known, but the intersecting node is unknown. Also, the number of nodes in each of the list before they intersect are unknown and both the list may have it different. List1 may have n nodes before it reaches intersection point and List2 might have m nodes before it reaches intersection point where m and n may be m = n, m > n or m < n. Give an algorithm for finding the merging point. Hints: A brute force approach would be to compare every pointer in one list with every pointer in another list. But in this case the complexity would be O(mn)Implement the three self-organizing list heuristics: Count – Whenever a record is accessed it may move toward the front of the list if its number of accesses becomes greater than the record(s) in front of it. Move-to-front – Whenever a record is accessed it is moved to the front of the list. This heuristic only works well with linked-lists; because, in arrays the cost of shifting all the records down one spot every time you move a record to the front is too expensive. Transpose – whenever the record is accessed swap it with the record immediately in front of it. Compare the cost of each heuristic by keeping track of the number of compares required when searching the list. Additional Instructions Use the SelfOrderedListADT abstract data type and the linked-list files I have provided to implement your self-ordered lists. You may incorporate the author’s linked list implementation via inheritance or composition, which ever makes the most sense to you (I will not evaluate that aspect of…The bounded-buffer solution in the below code uses a last-in-first-out strategy (LIFO).Change the code to implement a FIFO (First-in-First-out) strategy. You may use the (in,out) pointer method(using semaphores to test if the queue is full or empty shouldalleviate the problem of only using up N-1 locations) or implement a FIFO queue. Usethe correct counting semaphore implementation. C code below ******************************************************************************************************************************* #include <stdio.h>#include <stdlib.h>#include <pthread.h>#include <semaphore.h>#define SIZE 5#define NUMB_THREADS 6#define PRODUCER_LOOPS 2typedef int buffer_t;buffer_t buffer[SIZE];int buffer_index;pthread_mutex_t buffer_mutex;/* initially buffer will be empty. full_semwill be initialized to buffer SIZE, which meansSIZE number of producer threads can write to it.And empty_sem will be initialized to 0, so noconsumer can read from buffer until a…