Question

Hopefully someone knows the answer to this Java-Certification question:

public static void main(String[] args) {
 String[] sa = {"d", "c", "a", "b" };
 int x = Arrays.binarySearch(sa, "b");
 Arrays.sort(sa);
 int y = Arrays.binarySearch(sa, "b");
 System.out.println(x + " " + y);
}

Which two results are possible? (Choose two.)
A) 7 0
B) 7 1
C) 7 3
D) -1 0
E) -1 1
F) -1 3

The only true answer is E) -1 1, because if you play through the binary-search-algorithm this is the only possible output. But they want me to choose two…
So the second one have to be B) 7 1 then, because the second binarySearch in the sorted array will always return 1.

So my Question is, why is B) 7 1 a possible result?
More specific: How is it possible, that the first binarySearch in the unsorted array returns 7?

Thanks in advance

Answer 1

This is a trick question. The results of binary search on an unsorted array are undefined, as per documentation:

The array must be sorted (as by the sort method, above) prior to making this call. If it is not sorted, the results are undefined.

This means in particular that any number, including seven, is fair game.

The results on a sorted one are well-defined: you’ll get a 1. As if by magic, there are only two pairs that end in 1 on the list of answers, so B and E is the choice the examiners expect you to make.