Question: Consider the following method, which implements a recursive binary search. /** Returns an index in arr where the value str appears if str appears *
Consider the following method, which implements a recursive binary search.
/** Returns an index in arr where the value str appears if str appears
* in arr between arr[left] and arr[right], inclusive;
* otherwise returns -1.
* Precondition: arr is sorted in ascending order.
* left >= 0, right < arr.length, arr.length > 0
*/
public static int bSearch(String[] arr, int left, int right, String str)
{
if (right >= left)
{
int mid = (left + right) / 2;
if (arr[mid].equals(str))
{
return mid;
}
else if (arr[mid].compareTo(str) > 0)
{
return bSearch(arr, left, mid - 1, str);
}
else
{
System.out.println("right");
return bSearch(arr, mid + 1, right, str);
}
}
return -1;
}
The following code segment appears in a method in the same class as bSearch.
String[] words = {"arc", "bat", "cat", "dog",
"egg", "fit", "gap", "hat"};
int index = bSearch(words, 0, words.length - 1, "hat");
How many times will "right" be printed when the code segment is executed?
A. 1
B. 2
C. 3
D. 7
E. 8
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