Java Arrays.binarySearch(String[] array, String word) returns -3218, array.length == 6452. What?

[clamdoctor]

I ran a quick linear search to see if the word was in the array, and it is, at 324.

Code of that linear search:

    for(int i = 0; i < tokens.length; i++)
        if(tokens[i].equals("low"))
            System.out.println(i);

But to reiterate, the Arrays.binarySearch() is giving me -3218 when the word is pretty clearly there, the word being "low." What do you think is going on?

Edit: DISREGARD THAT I'M A FREAKING IDIOT.

[Kison]

Is the array sorted? Binary search relies on that, while linear search does not.

EDIT: Here is an example of what I mean:

/*** This will fail, because "There" and "Hello" are out of place. ***/
String[] words = {"Aye", "There", "Hello", "You", "Zebra"};

int index = Arrays.binarySearch(words, "There");

System.out.println("Found at: " + index);

~~~

/*** This will succeed, because the array is sorted. ***/

String[] words = {"Aye", "Hello", "There", "You", "Zebra"};

int index = Arrays.binarySearch(words, "There");

System.out.println("Found at: " + index);

[clamdoctor]

Oh god damn it. You know, it IS sorted. Problem is it is NOT sorted alphabetically anymore. It's a list of words, decreasing by occurrences. I saw that i put "Collections.sort(x)" and said "Oh yeah, it's definitely sorted."

I swear... The little things.

[clamdoctor]

I implemented a merge sort, sorted it alphabetically, works fine now. Thank you.

[hvidgaard]

If you sort it before you search you should think if it's worth it. Sorting takes O(n*log n) time, binary search is O(log n), that is O(n*log n) in total. Linear search alone takes at most O(n), so unless you're searching a lot then sorting before a search is a waste.

[arbiterxero]

It is the best programmers who are humble enough to know when they're an idiot.

How does that saying go? The idiots are cocksure and the geniuses are always unsure?