Set: Set is a data structure that is used for storing and processing non duplicate elements. It is an interface which extends collection and it is an unordered collection of objects. Set is implemented by LinkedHashSet, HashSet or a TreeSet. LinkedHash Set: In order to store its elements internally, the LinkedHashSet uses Linked HashMaps and it is an ordered version of HashSet that maintains a doubly-linked List across all elements. Hash Set: In order to store its elements internally, the HashSet uses HashMaps and it implements the Set interface, backed by a hash table which is actually a HashMap instance

Question

Want to see the full answer?

Answer 1

Question

Chapter 21.4, Problem 21.4.1CP

Program Plan Intro

Set:

Set is a data structure that is used for storing and processing non duplicate elements. It is an interface which extends collection and it is an unordered collection of objects.

Set is implemented by LinkedHashSet, HashSet or a TreeSet.

LinkedHash Set:

In order to store its elements internally, the LinkedHashSet uses Linked HashMaps and it is an ordered version of HashSet that maintains a doubly-linked List across all elements.

Hash Set:

In order to store its elements internally, the HashSet uses HashMaps and it implements the Set interface, backed by a hash table which is actually a HashMap instance

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

Contact list: Binary Search A contact list is a place where you can store a specific contact with other associated information such as a phone number, email address, birthday, etc. Write a program that first takes as input an integer N that represents the number of word pairs in the list to follow. Word pairs consist of a name and a phone number (both strings). That list is followed by a name, and your program should output the phone number associated with that name. Define and call the following function. The return value of FindContact is the index of the contact with the provided contact name. If the name is not found, the function should return -1 This function should use binary search. Modify the algorithm to output the count of how many comparisons using == with the contactName were performed during the search, before it returns the index (or -1). int FindContact(ContactInfo contacts[], int size, string contactName) Ex: If the input is: 3 Frank 867-5309 Joe 123-5432 Linda…

in cpp please AnyList.h #ifndef ANYLIST_H#define ANYLIST_H #include<iostream>#include <string> //Need to include for NULL class Node{public: Node() : data(0), ptrToNext(nullptr) {} Node(int theData, Node *newPtrToNext) : data(theData), ptrToNext(newPtrToNext){} Node* getPtrToNext() const { return ptrToNext; } int getData( ) const { return data; }void setData(int theData) { data = theData; } void setPtrToNext(Node *newPtrToNext) { ptrToNext = newPtrToNext; } ~Node(){}private:int data; Node *ptrToNext; //pointer that points to next node}; class AnyList{public: AnyList(); void insertFront(int); void print() const; void destroyList(); ~AnyList(); private: Node *ptrToFirst; //pointer to point to the first node in the list int count; //keeps track of number of nodes in the list}; #endif AnyList.cpp #include "AnyList.h" using namespace std; //constructorAnyList::AnyList(){ ptrToFirst = nullptr; count =…

#include <stdio.h> int arrC[10] = {0}; int bSearch(int arr[], int l, int h, int key); int *joinArray(int arrA[], int arrB[]) { int j = 0; if ((arrB[0] + arrB[4]) % 5 == 0) { arrB[0] = 0; arrB[4] = 0; } for (int i = 0; i < 5; i++) { arrC[j++] = arrA[i]; if (arrB[i] == 0 || (bSearch(arrA, 0, 5, arrB[i]) != -1)) { continue; } else arrC[j++] = arrB[i]; } for (int i = 0; i < j; i++) { int temp; for (int k = i + 1; k < j; k++) { if (arrC[i] > arrC[k]) { temp = arrC[i]; arrC[i] = arrC[k]; arrC[k] = temp; } } } for (int i = 0; i < j; i++) { printf("%d ", arrC[i]); } return arrC; } int bSearch(int arr[], int l, int h, int key) { if (h >= l) { int mid = l + (h - l) / 2; if…