BINARY SEARCH

The binary search program can be used it's only a sorted list of elements that means the binary search is used only with the list of elements that already arrange in order

The binary search cannot be used for a list of elements in a random order this process starts comparing the search element if the middle element in the list both are Matched then the result is element found 

otherwise we need to check whether the search element  is smaller or larger than the middle element in the list if the search element smaller then repeat the same process for the left sublist of the middle element if the search element is larger then we repeat the same process for the rights of sublist of the middle element 

repeat this process until we find the set element in the list or until left with sub list of only element and if that element also doesn't match with the search element then the result it as element not found in the list

Let us taken an example

A[5]      =         [79]. [26]. [88]. [11]. [65]

                         A[0]  a[1]  a[2]   a[3]  a[4]

By using bubble sort technique the sorted array

A[5]      =         [11]. [26]. [65]. [79]. [88]

                         A[0]  a[1]  a[2]   a[3]  a[4]

Let 79 search element

Start=0,end=4

Mid=0+4/2=2

IF a[mid]=search element then the result is element found but here a(2)=63=/79(search element)

Now we have to check whether the search element is less than or greater than a[mid]

65<79

A[mid] search element

Now it will repeat the process on the right sub array

Start=mid+1 end=4

=2+1

=3

Mid=start+end/2=3+4/2=7/2=3.5=a(3)=79

Here the search element is equal to the a[mid] of right sub array hence the result will be element 

PROGRAM ON BINARY SEARCH:

import java.util.Scanner;

class binarysearch
 {
  public static void main(String[] args)
  {
   int i,j,start,end,mid,n,item,temp,flag=0;
   Scanner bs = new Scanner(System.in);
   System.out.println("Enter Size Of Array :");
   n=bs.nextInt();
   int[] a;//declaration of array
   a=new int[n]; // Creation of array with given size
   System.out.println("Enter unsorted Array :");
   for(i=0;i<n;i++)
    {
     System.out.printf("\n Element a[%d]  :",i);
     a[i]=bs.nextInt();
    }
  
   for(i=0;i<n-1;i++)
    {
     for(j=0;j<n-1-i;j++)
      {
        if(a[j]>a[j+1])
        {
         temp=a[j];
         a[j]=a[j+1];
         a[j+1]=temp;
        }
      }
    }
   System.out.print("\n Sorted Array elements :");
   for(i=0;i<n;i++)
     {
      System.out.print( a[i] +" ");
     }

   for(i=0;i<n;i++)
    {
     System.out.printf("\n ADDRESS=%h ---- ARRAY=a[%d]----- DATA=%d --- location=%d\n",i*4,i,a[i],i+1);
    }
   System.out.print("\n Enter the element to be searched :");
   item=bs.nextInt();
   start=0;end=n-1;
      while(start <= end)
       {
        mid=(start+end)/2;
        if(item < a[mid])
          end=mid - 1;
        else if(item > a[mid])
          start=mid + 1;
        else if(item == a[mid])
          {
           System.out.printf("\n Element %d is found at location %d",item,mid+1);
           flag=1;
           break;
          }
       }
   
   if (flag==0)
      System.out.printf("\n Element %d is not in the array",item);
 }
}

output: