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Consider the following method, which implements a recursive binary search. /** Returns an index in theList where target appears,
* if target appears in theList between the elements at indices
* low and high, inclusive; otherwise returns -1.
* Precondition: theList is sorted in ascending order.
* low >= 0, high < theList. size(), theList. size() > 0
*/
public static int binarySearch(ArrayList theList, int low, int high,
int target)
{
if (low > high)
{
return -1;
}
int middle = (low + high) / 2;
if (target == theList. get(middle))
{
return middle;
}
else if (target < theList. get(middle))
{
return binarySearch(theList, low, middle - 1, target);
}
else
{
return binarySearch(theList, middle + 1, high, target);
}
}
The following code segment appears in a method in the same class as binarySearch.

ArrayList theList = new ArrayList ();
for (int k = 10; k < 65; k = k + 5)
{
theList. add(k);
}
int result = binarySearch(theList, 0, theList. size() - 1, 45);
Including the call to binarySearch in the last statement of the given code segment, how many times will binarySearch be called before a value is returned?

1- A

2- B

3- C

4- D

8- E

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Answers: 1

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Consider the following method, which implements a recursive binary search. /** Returns an index in...
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