Page 336 - ComputerScience_Class_11

Now, 8 > ar[2], i.e., 6.
So, it has been concluded that the element to be searched is in the right half of the array.
Now, min = mid + 1 = 2 + 1 = 3
Step 2: Repeat the above step.

Position or index 0 1 2 3 4
Ar 1 4 6 8 10
min Max
mid
Middle element is 8 as (3 + 4) / 2 = 7 / 2 = 2 (taking only the integer part)

Now, 8 == ar[3], i.e., 8.
Thus, the searched element is found and the loop breaks.
Since the array is already sorted, the binary search technique is faster than the linear search technique.

Program 9 Write a Java program to input “n” names in an array and a name to search for and print
whether the name is present in the array or not. (Use the binary search technique)
1 import java.util.*;

2 class Binary_search
3 {

4 public static void main(String args[])
5 {

6 Scanner sc= new Scanner(System.in);
7 System.out.print("How many names you want to enter? ");

8 int n = sc.nextInt();
9 String ar[]=new String[n];

10 int i, pos=-1, min=0, mid=0, max=n;
11 String ns;

12 for (i=0; i<n; i++)
13 {
14 System.out.print("Enter a name: ");

15 ar[i] =sc.next();

16 }
17 System.out.print("Enter a name to search: ");
18 ns=sc.next();

19 while(min<max)
20 {

21 mid=(min+max)/2;
22 if(ar[mid].equalsIgnoreCase(ns))

334 Touchpad Computer Science (Ver. 3.0)-XI