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    <p>I don't see why you'd need an array of hashsets. Here's what I invision:</p> <pre><code>Set&lt;String&gt; dictionary = new HashSet&lt;String&gt;(); dictionary.add("One"); dictionary.add("Two"); dictionary.add("Three"); dictionary.add("Four"); </code></pre> <p>And here is how I would use it. Note: don't read below unless you want an actual answer to the breaking-into-words problem. It may diminish how much learning you get. So only read it if you're okay with it being spoiled.</p> <pre><code>List&lt;String&gt; split(String sentence) { List&lt;String&gt; words = new LinkedList&lt;String&gt;(); String word = ""; // StringBuilder actually is not orders faster in // this case or I would advocate using it... for(int i = 0; i &lt; sentence.length(); i++) { word += sentence.charAt(i); // creates a new String anyway, so StringBuilder // is far less powerful if(dictionary.contains(word) { words.add(word); word = ""; } } return words; } </code></pre> <p>Some concerns:</p> <p>Let us assume your sentences and words are all in lowercase, to avoid case sensitivity. Let us also assume your dictionary contains every common English word.</p> <pre><code>dictionary.add("this"); dictionary.add("is"); dictionary.add("a"); dictionary.add("test"); </code></pre> <p>And run "thisisatest" and it will split it properly.</p> <p>Now, keep in mind there are other words.</p> <pre><code>dictionary.add("i"); dictionary.add("sat"); dictionary.add("est"); </code></pre> <p>These are all valid words. Running it will give you</p> <pre><code>"this" "i" "sat" "est" </code></pre> <p>In fact, by this logic, EVERY word that starts with <code>i</code> or <code>a</code> will end up getting missed. And that's bad. Especially for words like <code>"apple"</code> You will get <code>a</code> as the first word, then continue searching for <code>"pple"</code> and words that start with <code>"pple"</code>. That's going to cause a lot of problems!</p> <p>Even if you can get around this problem, you're going to run into problems where the words are always valid.</p> <p>Consider "thetreescare". Is it "the" "tree" "scare" or "the" "trees" "care". You are unable to make the differentiation - ever!</p> <p>So the problem you have picked is a dousie, for sure!</p>
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    1. POPicking the right data structure
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    1. COBut would StringBuilder() not produce less references? "word +=" creates a new String object.
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    2. CO@user968951 The problem is that you would need to create that String object anyway in order to check the map. Although, now that I look at it, you have to create StringBuilder objects for the +=... so it is faster in that sense. But it's not orders faster which StringBuilder intends. Also, if I specifically use a StringBuilder, I will miss out on optimizations in future versions of the JDK. (Consider people who had StringBuffer and had to change it to StringBuilder...)
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    3. COThanks for the quick response. For this I guess I would have to go through the whole /usr/share/dict/words and call dictionary.add on each one, then iterate split every time it finds a word. Seems like it would work.
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      1. USDarth Egregious
 

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