This Is In Java Exercise 10.2.8: Maximum Iterations 5 Points Let's Go! How Many Times Does A Binary Search Need To Execute To Find Its Value? Recall From Our Lesson That The Number Of Iterations Is Roughly A Log Base 2 Relationship. In This Exercise, You Are Going To Calculate The Maximum Iterations And The Actual Iterations Needed To Find A Random Value InThis is in javaExercise 10.2.8: Maximum Iterations5 pointsLet's Go!How many times does a binary search need to execute to find its value? Recall from our lesson that the number of iterations is roughly a log base 2 relationship.In this exercise, you are going to calculate the maximum iterations and the actual iterations needed to find a random value in arrays of size 100, 1000, 10k, and 100k.You are given helper methods to calculate the maximum iterations, generate the random array, and do the binary search. You are also given a counter variable that increments up each time your recursive binary method is called.You will need to come up with the remainder of the code. Same output is provided below.Sample OutputArray Size: 100Max iterations: 7Actual iterations: 5Array Size: 1000Max iterations: 10Actual iterations: 5BinarySearchTest.javaimport java.util.*;public class BinarySearchTest {static int count;public static void main(String[] args) {// Use the helper code to generate arrays, calculate the max// iterations, and then find the actual iterations for a randomly// selected value.}public static int binaryRec(int[] array, int target, int begin, int end) {if (begin <= end){int mid = (begin + end) / 2;count ++;// Base Caseif (target == array[mid]) {return mid;}if (target < array[mid]) {return binaryRec(array, target, begin, mid - 1);}if (target > array[mid]) {return binaryRec(array, target, mid + 1, end);}}return -1; //Alternate Base Case - not found}public static int[] generateArrayOfLength(int length){int[] arr = new int[length];for(int i = 0; i < length; i++){arr[i] = (int)(Math.random() * 100);}Arrays.sort(arr);return arr;}private static int binaryMax(int length) {return (int) (Math.log(length) / Math.log(2)) + 1;}}
