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

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
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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
Transcribed Image Text: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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