String[] words = {"one", "two", "tree", "four", "five"};

String[] words2 = {"blah", "blih", "bluh"};

String stringToSearch1 = "My garden has a tree";

String stringToSearch2 = "My house bluh";

ArrayList <String> one = ...;

ArrayList <String> two = ... ;

Now i need to add the strings to different ArrayList based in matching one or the other. If the string contains words form words array goes to one ArrayList, if contains words from words2 goes to the other. The strings to be compared also are an array i'm traversing one by one...

like String[] toBecompared = loadStrings(path);

  

vk

 The first code, from Sergio, is easier for me to understand, but it produces an odd result in the last comparison:

There is no "use" in "five empty houses".

Benja's code worked fine, but i need some help to understand it. Questions:

1- The noLoop() at the end of setup makes draw() run only once? I would have placed it at the end of draw...

2-  ArrayList<ArrayList<String>> sortedStrings; well  this i've never seen before, so it is an arraylist of arraylists of strings?  The cleverness here is that this way, as in words[][], we can interact with both at the same loop, right? Good idea!

But, this:

Is it just adding empty (null?) elements to the array list? If so, why is this necessary? I mean an arraylist can grow as needed right?

3 The access seems weird, i mean i need to use get() even to add(). Correct?  sortedStrings.get(j).add(toBeCompared[i]);

      so to reproduce the line 2darray[100][88] = 1000; i need to do  2darraylist.get(100).get(88) =1000;  ?

Any way that is working!! Thanks again, gonna implement it on the sketch now :-)

A last question, in this post, answering me, PhiLho sad that a hashMap is faster for comparison. I don't think i would need the fastest way, but would that be an approach fro this matter? 

Thanks you all.

vk

There is a "use" in "five empty ho USEs".
To avoid these results, i think you could split the strings into words, and store it in a HashMap, for comparing words, that would be the fastest solution i guess.

This creates an empty ArrayList, wich can hold ArrayLists of type String. But there are no ArrayLists at this point, so we have to create some, that is what happens in the loop: We add one ArrayList<String> to sortedStrings, for each word-list.

3. Yes the syntax is different, you have to use get() to access an element of an ArrayList. Might look more complicated, but you will get used to it.

and for 1:

I get it now I need arraylists to put on an arraylist of arraylists ;-)

Thanks;