No hand written and fast answer with explanation Here is the Java code for a recursive binary search:Coding ExerciseRun the code below. Try searching for the value 3 and then the value 2 which is not in the array. What would happen if we removedthe second base case checking if end < start? Try it and see. RecursiveBinarySearch.java12345678910111213141516171819202122232425262728public class RecursiveBinarySearch{public static int recursiveBinarySearch(int[] array, int start, int end, int target){29303132 }33int middle = (start + end)/2;// base case: check middle element}if (target == array[middle]) {return middle;}// base case: check if we've run out of elementsif(end < start) {return -1; // not found}// recursive call: search start to middleif (target < array[middle]){return recursiveBinarySearch(array, start, middle 1, target);}// recursive call: search middle to endif (target > array[middle]){return recursiveBinarySearch(array, middle + 1, end, target);}public static void main(String[] args)(}return -1;int[] array { 3, 7, 12, 19, 22, 25, 29, 30 };.int target 25;int foundIndex recursiveBinarySearch(array, e, array.length-1, target);System.out.println(target + was found at index"+foundIndex);

All the 3 questions related to the code are well explained below along with output screenshot

Coding Exercise Run the code below. Try searching for the value 3 and then the value 2 which is not in the array. What would happen if we removed the second base case checking if end < start? Try it and see.

Coding Exercise Run the code below. Try searching for the value 3 and then the value 2 which is not in the array. What would happen if we removed the second base case checking if end < start? Try it and see.

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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Question

No hand written and fast answer with explanation

Here is the Java code for a recursive binary search:
Coding Exercise
Run the code below. Try searching for the value 3 and then the value 2 which is not in the array. What would happen if we removed
the second base case checking if end < start? Try it and see.

RecursiveBinarySearch.java
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public class RecursiveBinarySearch
{
public static int recursiveBinarySearch(int[] array, int start, int end, int target)
{
29
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32 }
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int middle = (start + end)/2;
// base case: check middle element
}
if (target == array[middle]) {
return middle;
}
// base case: check if we've run out of elements
if(end < start) {
return -1; // not found
}
// recursive call: search start to middle
if (target < array[middle]){
return recursiveBinarySearch(array, start, middle 1, target);
}
// recursive call: search middle to end
if (target > array[middle]){
return recursiveBinarySearch(array, middle + 1, end, target);
}
public static void main(String[] args)
(
}
return -1;
int[] array { 3, 7, 12, 19, 22, 25, 29, 30 };
.
int target 25;
int foundIndex recursiveBinarySearch(array, e, array.length-1, target);
System.out.println(target + was found at index"+foundIndex);

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