Write a function that takes in n ADT Lists in an array. The function is going to be called Intersection, and it will return a new list with the elements common to all the n lists. Determine the time complexity
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- take & array from user and print all the elements after doubling its and you have to separate function for doubling operation.Rewrite the Binary search function in the textbook to use a generic type for the array elements. Test the function with arrays of int, double, and string values. *When using test arrays make sure elements are stored in ascending sorted order*You have been given two integer arrays/list(ARR1 and ARR2) of size N and M. respectively. You need to print their intersection; An intersection for this problem can be defined when both the arrays/lists contain a particular value or to put it in other words, when there is a common value that exists in both the arrays/lists. The first line contains an Integer Y which denotes the number of test cases or queries to be run Then the test cases follow. First line of each test case or query contains an integer 'N' representing the size of the first array/list. Second line contains 'N' single space separated integers representing the elements of the first the array/list. Third line contains an integer 'M' representing the size of the second array/list Fourth line contains 'M' single space separated integers representing the elements of the second array/list Output format For each test case print the intersection elements in a row, separated by a single space. Output for every test case will be…
- create an array of 30 random numbers that range between 1and 100. And yet again, write a function that will receive a number from the userand determine if that number exists in the array or not. But this time, start bySORTING your input list. After a sort, the list in problem 1 is as follows:[2, 2, 3, 5, 12, 14, 14, 15, 23, 36, 39, 41, 44, 44, 45, 48,49, 50, 52, 52, 59, 71, 81, 82, 88, 89, 89, 93, 96, 97] Approach: Implement a method called findC(x, A, i, j), where x is the number we arelooking for in array A, the first index of the array is i and the last index is j. We wantto determine whether x exists in A anywhere between index i and index j. Your firstcall to this method will therefore look like this: findC(x, A, 0, A.length-1). In the body of your function, compare x with the item that is in the middle of thearray, as you did before. As before, call the middle of index of the array mid. But thistime, if x<=a[mid], recursively call your function to search ONLY the first half of…The function interleave_lists in python takes two parameters, L1 and L2, both lists. Notice that the lists may have different lengths. The function accumulates a new list by appending alternating items from L1 and L2 until one list has been exhausted. The remaining items from the other list are then appended to the end of the new list, and the new list is returned. For example, if L1 = ["hop", "skip", "jump", "rest"] and L2 = ["up", "down"], then the function would return the list: ["hop", "up", "skip", "down", "jump", "rest"]. HINT: Python has a built-in function min() which is helpful here. Initialize accumulator variable newlist to be an empty list Set min_length = min(len(L1), len(L2)), the smaller of the two list lengths Use a for loop to iterate k over range(min_length) to do the first part of this function's work. On each iteration, append to newlist the item from index k in L1, and then append the item from index k in L2 (two appends on each iteration). AFTER the loop…* allSame returns true if all of the elements in list have the same value. * allSame returns false if any two elements in list have different values. * The array may be empty and it may contain duplicate values. * * Your solution should contain at most one loop. You may not use recursion. * Your solution must not call any other functions. * Here are some examples (using "==" informally): * * * * * * * * true == allSame (new double[] { }) true == allSame(new double[] {11}) true == allSame (new double[] { 11, 11, 11, 11 }) false == allSame(new double[] { 11, 11, 11, 22 }) false == allSame (new double[] { 11, 11, 22, 11 }) true == allSame (new double[] { 22, 22, 22, 22 }) * */ public static boolean allSame (double[] list) { return StdRandom.bernoulli(); //TODO: fix this
- Write a function that gets an array as a parameter, sorts it in an ascending fashion and returns the number of elements that did not change position in the sort (they were originally in the right position). Use c code int sortASC(int* myArray, int size) if array ={ 2, 3, 1, 0, -1} after calling the function the array should become {-1, 0, 1, 2 , 3 } and the function should return 1 (only one element has not changed place in the array)Write a function that takes a nested list of integers and add up the elements from all of the nested lists.You are given the pointer to the head node of a linked list and an integer to add to the list. Create a new node with the given integer. Insert this node at the tail of the linked list and return the head node of the linked list formed after inserting this new node. The given head pointer may be null, meaning that the initial list is empty. Function Description Complete the insertNodeAtTail function in the editor below. insertNodeAtTail has the following parameters: SinglyLinkedListNode pointer head: a reference to the head of a list int data: the data value for the node to insert Returns SinglyLinkedListNode pointer: reference to the head of the modified linked list Input Format The first line contains an integer , the number of elements in the linked list.The next lines contain an integer each, the value that needs to be inserted at tail. Constraints Sample Input STDIN Function ----- -------- 5 size of linked list n = 5 141 linked list data values…
- Write a function with below starting function: public static void printFirstAnd Last(ArrayList list) The approach should accomplish the following: Print "no elements" if the list is empty. Print the single element in the list if there is only one. Print the first entry in a list with more than one element, followed by a space, and then the last element.In main() for now: do these - one at a time, each in it's own loop (we will make functions out of them later –Declare an array RandArray with 20 integers –Assign each of the 20 values with a random value from 0 to 99 Hint: Use rand()%100 For mimir: Do not call srand at the top of main. // normally: Call srand(time(0)) at the top of main –(you need #include<cstdlib>) –Write another loop that prints the array, with index values Important: If the output values do not match mimir, please add srand(17); // inside your main function - at the top code format: #include <iostream>using namespace std; #include <cstdlib> // required for rand() int main(){ srand(17); // define a constant ARRAYSIZE that is 20 // declare randArray // set the 20 elements in randArray to be a random number between 0 and 99 // hint: use rand()%100 // print the 20 values return 0;}Write a function is ordered that takes an integer array and its size as parameters and returns true if the array elements are ordered in nondecreasing order and returns false if not. For example, integers in {10,0,1,1,5,1009} are ordered in nondecreasing order. On the other hand, integers in {1, 2, 3, 8, 4, 19} are not since 8 > 4