Question

Given the following code for a binary search, how many times this method have to be executed in order to find the number 5?

A) 1 time

B) 2 times

C) 3 times

D) 4 times

E) more than 5 times

public class BinarySearch{

public static void main(String []args){
if (right >= left) {

int middle = left + (right - left) / 2;

if (arr[middle] == num)

return middle;

if (arr[mid] > num)

return binarySearch(arr, left, middle - 1, num);

return binarySearch(arr, middle + 1, right, num);

}

return -1;
  
}

int[] arr = {1, 3, 5, 6, 7, 9, 11, 15, 18}

return binarySearch(arr, 0, 8, 5)

int binarySearch(int arr[], int left, int right, int num)

{

  

}
}

  

Answer 1

public class BinarySearch {

    static int count = 0;

    public static void main(String[] args) {
        int[] arr = {1, 3, 5, 6, 7, 9, 11, 15, 18};
        System.out.println(binarySearch(arr, 0, 8, 5));
        System.out.println("Number of times binarySearch is executed is " + count);
    }

    static int binarySearch(int arr[], int left, int right, int num) {
        ++count;
        if (right >= left) {
            int middle = left + (right - left) / 2;
            if (arr[middle] == num)
                return middle;
            if (arr[middle] > num)
                return binarySearch(arr, left, middle - 1, num);
            return binarySearch(arr, middle + 1, right, num);
        }
        return -1;
    }
}

C) 3 times