In the following code, identify the first and second base case, and evaluate whether this code would have an efficient running time.
1 public static int recursiveBinarySearch
2 ( int [ ] arr, int key, int start, int end )
3 {
4 if ( start <= end )
5 {
6 int middle = ( start + end ) / 2;
7 if ( arr[middle] == key )
8 return middle;
9 else if ( arr[middle] > key )
10 return recursiveBinarySearch( arr, key, start, middle - 1 );
12 else
13 return recursiveBinarySearch( arr, key, middle + 1, end );
14 }
15 else
16 return -1;
17 }
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