java Search for the character Z using the binary search algorithm on thefollowing array of characters: ADHJLNPRZFor each iteration of binary search use a table similar to the tablebelow to list: (a) the left index and (b) the right index of the arraythat denote the region of the array that is still being searched, (c)the middle point of the region of the array that is still being searched, and (d) the character-to-character number of comparisons made during the search (line 09 and line 11 of the BinarySearch algorithm).01 public static int binarySearch (char [] a, char target) {0203int left = 0;04int right = a.length - 1;0506while ( left <= right {07int middle = ( left + right ) / 2;if ( a[middle]0809== target ) {10return middle;} else if ( target < a[middle] ) {11right = middle - 1;} else {121314left = middle + 1;15}16}return -1; // target not found18 }17IterationLeft | Right | Middle | Number of Comparisons13Enter your answer here

It is a collection of nodes arranged in Binary search tree properties. Basic operation of Binary…

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Search for the character Z using the binary search algorithm on the following array of characters: A DHJLNPRZ

Programming Logic & Design Comprehensive

9th Edition

ISBN:9781337669405

Author:FARRELL

Publisher:FARRELL

Chapter6: Arrays

Section: Chapter Questions

Problem 19RQ

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Question

java

Search for the character Z using the binary search algorithm on the
following array of characters: ADHJLNPRZ
For each iteration of binary search use a table similar to the table
below to list: (a) the left index and (b) the right index of the array
that denote the region of the array that is still being searched, (c)
the middle point of the region of the array that is still being searched, and (d) the character-to-
character number of comparisons made during the search (line 09 and line 11 of the Binary
Search algorithm).
01 public static int binarySearch (char [] a, char target) {
02
03
int left = 0;
04
int right = a.length - 1;
05
06
while ( left <= right {
07
int middle = ( left + right ) / 2;
if ( a[middle]
08
09
== target ) {
10
return middle;
} else if ( target < a[middle] ) {
11
right = middle - 1;
} else {
12
13
14
left = middle + 1;
15
}
16
}
return -1; // target not found
18 }
17
Iteration
Left | Right | Middle | Number of Comparisons
1
3
Enter your answer here

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