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Basic Binary Search

The middle element of the array is compared to the target value in a binary search. If they aren't equal, the method determines which side of the array the element is in (left or right) and repeats the same process until it finds the element. Get an array of Integers arr[] with size n and element e to be found. Apply Linear Search to find the element e in the array arr[]. If element e is found, print its index; if not, print -1.

Examples

Input 1:

5

5 7 9 3 1

9

Output 1:

2

Input 2:

5

5 7 9 3 1

4

Output 2:

-1

Try your Solution

Strongly recommended to Solve it on your own, Don't directly go to the solution given below.

public class Hello { public static void main(String args[]) { //Write your code here } }

Program or Solution

				
				
					
import java.util.*;
class SearchB
{
	public static void main(String args[])
    	{

		//Variable and  Object Declarations
       		Scanner input = new Scanner(System.in);
		int n,e,pos=-1;
		
		//Getting size of Array & Declare Array arr[]
		n = input.nextInt();
		int arr[] = new int[n];

		//Get n values to array arr[]
		for(int i = 0; i<n; i++)
		{
			arr[i] = input.nextInt();
		}

		e = input.nextInt();

	
		int left=0, right=n-1;
		
		//Binary Search
		while(left<=right)
		{
			int mid = (left+right)/2;
			if(arr[mid]==e)
			{
				pos = mid;
				break;
			}
			else if(arr[mid]>e)
			{
				right = mid-1;
			}
			else
			{
				left = mid+1;
			}
			
		}
	
		System.out.println(pos);
		
      	}
}

Program Explanation

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