Given the QuickSort implementation from class, provide an 18-element list that will take the least number of recursive calls of QuickSort to complete. As a counter-example, here is a list that will cause QuickSort to make the MOST number of recursive calls: public static List input() { return Accaxs.asbist(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0); } And here's the QuickSort algorithm, for convenience: algorithm QuickSant Input: lists of integers 1st of size N Output: new list with the elements of 1st in sorted order if N < 2 return st pivot = lst[N-1] left = new empty list right = new empty list for index = 0, 1, 2, if lat[i]<= pivot left.add(lat[i]) else ... N-2 cight.add(lst[i]) return Quicksact(left) + [pivot] + QuiskSort (right)

This program is about identifying palindromes. A palindrome is a list that reads the same forwardas backward. For example, [2, 5, 2] is a palindrome. [2, 5, 2, 2] is not a palindrome. Write a prologfunction, is_palindrome(X), that returns true if the list X is a palindrome, otherwise it returns false. Notethat an empty list is a palindrome.Examples:is_palindrome([1,1,3,1,1]) should return true.is_palindrome([1,3,1,1]) should return false.is_palindrome([1,3,3,1]) should return true.is_palindrome([a,b,l,e,w,a,s,i,e,r,e,i,s,a,w,e,l,b,a]) should return true.

Can you write a program in python for the dining philosopher problems that do not have deadlock? You have to use semaphores and use the strategy that a philosopher only picks the fork if both (left and right) are available. def philosopher(id: int, fork: list): The main is if __name__ == "__main__": semaphoreList = list() #this list will hold one semaphore per fork Philosophers = 5 for i in range(Philosophers): semaphoreList.append(multiprocessing.Semaphore(1)) philosopherList = list() for i in range(Philosophers): #instantiate all processes representing philosophers philosopherList.append(multiprocessing.Process(target=philosopher, args=(i, semaphoreList))) for j in range(Philosophers): #start all child processes philosopherList[j].start() for k in range(Philosophers): #join all child processes philosopherList[k].join() sample output: DEBUG: philosopher2 has chopstick2 DEBUG: philosopher2 has chopstick3 DEBUG: philosopher2 eating DEBUG: philosopher1 has chopstick1 DEBUG: philosopher0…

Implement a function countMultiples in Python that: accepts a single argument, a 2-dimensional list (e.g, list of lists) containing non-negative integers, and returns the count of those integers that are multiples of 5 but that are not not multiples of 10. (e.g., 65 should be counted, but 80 should not be counted) sample usage: >>> countMultiples( [[1,2,3],[4,5],[6,7,8,9]] ) # 5 counts1