Enthuware

We have the array Why is the answer "Arrays.binarySearch(sa, "c", new MyStringComparator()); will return 0." correct? To my understanding, "c" does not exist and will be inserted at index 0, thus Arrays.binarySearch(sa, "c", new MyStringComparator()); will return -(0)-1 = -1. However, that answer was marked wrong.

Why do you think "c" will be inserted at index 0? The explanation tells exactly where it will be inserted: "The insertion point is defined as the point at which the key would be inserted into the list: the index of the first element greater than the key,..."

binarySearch would return 0 if "c" would be found in array at index 0, right?

The first element greater than "c" is "bbb" with index 1. Then binarySearch should return -(1+1) = -2.

<EDIT> The explanation says exactly that about return value 0: "Note that this guarantees that the return value will be >= 0 if and only if the key is found. >

That's the thing, as Elecded explained.

It might be still somehow confusing in my mind, but I believe the reason behind 0 as the search index for "c" key is that our comparator is based on length and not on the natural (alphabetical) order of elements. "d" is of the same length and the key that we search, so, our custom Comparator class returns 0

thodoris.bais wrote:That's the thing, as Elecded explained.

It might be still somehow confusing in my mind, but I believe the reason behind 0 as the search index for "c" key is that our comparator is based on length and not on the natural (alphabetical) order of elements. "d" is of the same length and the key that we search, so, our custom Comparator class returns 0

Your understanding is correct.

To my understanding only works if the array has been previously sorted. ("The array must be sorted (as by the method) prior to making this call." --> Otherwise)
The array is not sorted. Why is the result not "undefined" then?

Thank you!

The array needs to be sorted as per the comparator you are using to search. In this case, we are using MyComparator, which compares the length of the string and according to this comparator, the array is indeed sorted.

Thanks!

Arrays.binarySearch() method returns the index of the search key, if it is contained in the list.

Would explain why is correct.

The "key" in

returns the index of the search key,

is defined by the comparator. In our case, the "key" is actually meant to be .length(), which is equal to 1 as to "d".
If "d" is passed into the method, 0 is returned because the string of length that equals to 1 is contained in the list and is located at index 0. From the Comparator point of view, the array is actually like {1, 3, 4} just because or compare() method forces the it to compare elements of the { "d", "bbb", "aaaa" } array by length.
Trying passing "d" (length=1), "ddd" (length=3)and "dddd" (length=4)into the call would return 0, 1, 2 respectively, which agrees to the quote and my conclusion.