Use java and use proper code indentation. Exercise 2 (Binary Search)Assume a binary search is performed on the following array of integers:{1, 14, 15, 24, 55, 59, 73, 90, 94, 99}Trace through each iteration of the algorithm, writing the number that will be the middleelement and the left and right bounds (indexes), when searching for the number 73.Explanation of the Binary Search ConceptBinary Search is an algorithm for finding a target element in a sorted linear collection. Itassumes that there exists a comparison relation between all elements in the collection thatallows for a total ordering of the elements. For example, integers can be compared by the lessthan (<) operator. For Strings, we can order them lexicographically and compare using thecompareTo method. The key assumption is that the linear collection is already sorted beforebinary search is applied.The binary search algorithm is naturally defined recursively as follows:if collection is empty, return -1 to indicate target is not foundCheck if middle element is the targetif yes, then return its indexelsé:if middle element is < target,then return outcome of binary search on the elements to the right of the middleelse return outcome of binary search on the elements to the left of the middle

The recursive version of binary search algorithm makes it difficult and hard to understand as to how to print middle value at each iteration and lower and upper bounds. Iterative method of binary search is efficient and tracing each iteration is convenient. Hence for the purpose of better understanding, iterative method of Binary search has been given.Source code: public class BinarySearch { public static void main(String[] args) { //declare the srray int arr[]= {1,14,15,24,55,59,73,90,94,99}; //declare and initialize left and right bound and middle int left=0, right=arr.length-1; int mid=left+right/2; //while left bound is less than or equal to right bound while(left<=right) { //calculate middle mid=(left+right)/2; //if middle element is equal to 73 then element found if(arr[mid]==73) { System.out.println("Middle element is "+arr[mid]); System.out.println("73 found"); System.out.println("Left bound= "+left+" Right bound= "+right); break; } //if 73 is less than middle element if(73<arr[mid]) { System.out.println("Middle element is "+arr[mid]); System.out.println("Left bound= "+left+" Right bound= "+right); System.out.println(); right=mid-1; } //else 73 is grater than middle element else { System.out.println("Middle element is "+arr[mid]); System.out.println("Left bound= "+left+" Right bound= "+right); System.out.println(); left=mid+1; } } } }Code screenshot:Output:

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Exercise 2 (Binary Search) Assume a binary search is performed on the following array of integers: {1, 14, 15, 24, 55, 59, 73, 90, 94, 99} Trace through each iteration of the algorithm, writing the number that will be the middle element and the left and right bounds (indexes), when searching for the number 73.

Exercise 2 (Binary Search) Assume a binary search is performed on the following array of integers: {1, 14, 15, 24, 55, 59, 73, 90, 94, 99} Trace through each iteration of the algorithm, writing the number that will be the middle element and the left and right bounds (indexes), when searching for the number 73.

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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