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- bool add(int anInt)// Pre: contains(anInt) ? size() <= MAX_SIZE : size() < MAX_SIZE// Post: If contains(anInt) returns false, anInt has been// added to the invoking IntSet as a new element and// true is returned, otherwise the invoking IntSet is// unchanged and false is returned.// bool remove(int anInt)// Pre: (none)// Post: If contains(anInt) returns true, anInt has been// removed from the invoking IntSet and true is// returned, otherwise the invoking IntSet is unchanged// and false is returned. bool add(int anInt); bool remove(int anInt); bool IntSet::add(int anInt){ if(used <= data[MAX_SIZE]) // check if the array is not full { return contains(used); //if not full then return to data } return false;} bool IntSet::remove(int anInt){ cout << "remove() is not implemented yet..." << endl; return false; // dummy value returned}Please convert the code to C language // C++ program to find minimum deletion // to make all elements same#include <bits/stdc++.h>using namespace std; int minimum_no_of_deletion(vector<int> arr, int n){ //create a hash table where for each key //the hash function is h(x)=x //we will use stl map as hash table and //will keep frequencies stored //so say two keys are mapping to the same location, //then the location will have value 2 //instead of the keys itself map<int, int> hash; for (auto i : arr) { //for each number hash[i]++; } //now to make all elements same //we need to keep only the keys with //maximum frequency //we need to delete the other keys //so find the key with max frequency int maxfreq = 0; for (auto it = hash.begin(); it != hash.end(); it++) { if (it->second > maxfreq) { maxfreq = it->second; } } //so we need to dlete rest of //the elements…please convert the code to C language // C++ program to find minimum deletion // to make all elements same#include <bits/stdc++.h>using namespace std; int minimum_no_of_deletion(vector<int> arr, int n){ //create a hash table where for each key //the hash function is h(x)=x //we will use stl map as hash table and //will keep frequencies stored //so say two keys are mapping to the same location, //then the location will have value 2 //instead of the keys itself map<int, int> hash; for (auto i : arr) { //for each number hash[i]++; } //now to make all elements same //we need to keep only the keys with //maximum frequency //we need to delete the other keys //so find the key with max frequency int maxfreq = 0; for (auto it = hash.begin(); it != hash.end(); it++) { if (it->second > maxfreq) { maxfreq = it->second; } } //so we need to dlete rest of //the elements…
- please update generateRandoms, letterToGrades, sumBetween, countCharacters //A6Utilities.java import java.io.IOException;import java.nio.file.Files;import java.nio.file.Paths;import java.security.SecureRandom;import java.util.List;import java.util.TreeMap;import java.util.function.Function;import java.util.stream.Collectors;public class A6Utilities {/*** Returns the corresponding letter grade. Don't modify this method, simply use it when converting single* grades to their equivalent letter grade** @param grade the integer grade to convert* @return the letter grade equivalent to the integer grade*/private static char letterGrade(final int grade) {if (grade < 0) return '?';else if (grade < 60) return 'F';else if (grade < 70) return 'D';else if (grade < 80) return 'C';else if (grade < 90) return 'B';else if (grade <= 100) return 'A';else return '?';}/*** The method uses Java Streams only. Don't use any looping structures, conditional statements are OK (IF, TERNARY,…Implement the function below void swap(int pos1, int pos2) {} For LinkedList, to maintain integrity of data in the structure, you are not to swap directly the element, nor remove a node. Instead, you are to only change the nodes' next pointers. Inital code to be completed: class LinkedList : public List { node* head; node* tail; int index; node* create_node(int num) { node* n = (node*) malloc(sizeof(node)); n->element = num; n->next = NULL; return n; } public: LinkedList() { index = 0; head = NULL; tail = NULL; } int add(int num) { addTail(num); return index; } int get(int pos) { node* currnode = head; int count =0; while (currnode != NULL){ count++; if (count == pos){ return currnode -> element; }else{ currnode = currnode -> next; } }…def invert_dict(d): """ Given a dictionary d, create the invert of the dictionary where the key is a value v in d, and v is mapped to a set of keys in original d whose values are equal to v. Return the inverted dictionary mapping value to set of keys """
- def max_kv(d): """ Given a dictionary, return a tuple of key value pair where the key is the largest in d. """Write a Python code using the given function and conditions. Do not use Numpy. Use LinkedList Manipulation. Given function: def insert(self, newElement, index) Pre-condition: The list is not empty. Post-condition: This method inserts newElement at the given index of the list. If an element with the same key as newElement value already exists in the list, then it concludes the key already exists and does not insert the key. [You must also check the validity of the index].Given a list mylist construct a dictionary mydict, the key is the index of an element in my list and its value will be the element value of mylist. for example the index of "a" is "0" . therefore the pair ( 0,"a" )will be in mydict .it means mydict = { 0:'a' ,1:'cybersecurity ', 2 : 'newyork'......} Do this in coding but not manually. given an empty dictionary mydict ={} write a python code to construct mydict for any given list. Hint : Your Code begins with mydict={}for val in mylist:
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