Exercise_1.java

class BinarySearch { 
    // Returns index of x if it is present in arr[l.. r], else return -1 
    int binarySearch(int arr[], int l, int r, int x) 
    { 
        //Write your code here
        if(l <= r)  //if element is not present right index will become greater than left index
        {
            int mid = l+(r-l)/2;
            if(x == arr[mid]){
                return mid;
            }
            else{
                if(x < arr[mid]){
                    return binarySearch(arr,l,mid-1,x);
                }else{
                    return binarySearch(arr,mid+1,r,x);
                }
            }
        }
        return -1;
    } 
  
    // Driver method to test above 
    public static void main(String args[]) 
    { 
        BinarySearch ob = new BinarySearch(); 
        int arr[] = { 2, 3, 4, 10, 40}; 
        int n = arr.length; 
        int x = 40; 
        int result = ob.binarySearch(arr, 0, n - 1, x); 
        if (result == -1) 
            System.out.println("Element not present"); 
        else
            System.out.println("Element found at index " + result); 
    } 
}