Consider the following method, which implements a recursive binary search.

/** Returns an index in arr where target appears, if target appears

* in arr between arr[low] and arr[high], inclusive;

* otherwise, returns -1.

* Precondition: arr is sorted in ascending order.

* low >= 0, high < arr.length, arr.length > 0

*/

public static int binarySearch(int[] arr, int low, int high, int target)

{

if (low > high)

{

return -1;

}

int middle = (low + high) / 2;

if (target == arr[middle])

{

return middle;

}

else if (target < arr[middle])

{

return binarySearch(arr, low, middle - 1, target);

}

else

{

return binarySearch(arr, middle + 1, high, target);

}

}

The following code segment appears in a method in the same class as binarySearch.

int[] arr = {2, 3, 12, 34, 54};

int result = binarySearch(arr, 0, arr.length - 1, 5);

If the first call to binarySearch is the call in the code segment above, with low = 0 and high = 4, which, if any, of the following shows the values of low and high when binarySearch is called for the third time?

A. low = 0, high = 1

B. low = 0, high = 2

C. low = 1, high = 1

D. low = 2, high = 1

E. The method returns to the calling code segment before the third call to binarySearch.

Question

ned72 · Accepted Answer

Answer:D. low = 2, high = 1Explanation:Middle is 2First call: low = 0 high = 1Middle is 0Second call: low = 1 high = 1Middle is 2Third call: low = 2 high = 1

uarmstrong · Answer

A technique of identifying a problem (or a set of strategies) in elements of (a simplified version of) itself is known as recursion. therefore, the final answer is "low = 1, high = 1".Recursion:In the given code, an integer array that is "arr" holds an integer value.In the next step, an integer variable "result" is declared that calls the binarySearch method.In the first time calls Recursion 1, when low = 0 and high = 4 its mid = 2, therefore arr[2] = 12 &gt; 5 = high = mid-1In the second time calls Recursion 2, when low = 0 and high = 1 its mid =0, therefore arr[0] = 2 &lt; 5 = low = mid+1In the third time calls Recursion 3, when low = 1 and high = 1 its mid =1, therefore arr[1] = 3 &lt; 5 = element not foundTherefore, the final answer is "Option C", that is "low = 1, high = 1".Find out more information about the Recursion here:brainly.com/question/17298888

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