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    <p><em>There's no need to make a <code>HashTable/HashMap/HashSet</code> for this.</em></p> <p><strong>You know which characters are being tracked ahead of time, so you can use an array.</strong></p> <blockquote> <p>I want to count the occurrence of all letters and numbers</p> </blockquote> <p>Make a string of the characters you will track, then initialize an array.</p> <pre><code>String sentence = "ABC ABC ABC 123"; //Make a map of all the characters you want to track. String indexes = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"; //Initialize an array to the size of the possible matches. int[] count = new int[indexes.length()]; //Loop through the sentence looking for matches. for (int i = 0; i &lt; sentence.length(); i++) { //This will get the index in the array, if it's a character we are tracking int index = indexes.indexOf(sentence.charAt(i)); //If it's not a character we are tracking, indexOf returns -1, so skip those. if (index &lt; 0) continue; count[index]++; } </code></pre> <p>Then you can print them all out with this:</p> <pre><code>for (int i = 0; i &lt; count.length; i++) { if (count[i] &lt; 1) continue; System.out.println(String.format("%s (%d) %s", indexes.charAt(i), count[i], //This little bit of magic creates a string of nul bytes, then replaces it with asterisks. new String(new char[count[i]]).replace('\0', '*'))); } </code></pre> <p>If you aren't comfortable with the <code>new String(new char[count[i]]).replace('\0', '*'))</code> bit, then you can use a <code>StringBuilder</code> to build the asterisk <code>String</code> before trying to output it. You can see <strong><em>@mike</em></strong>'s example below for a good example of that.</p> <h3>Outputs</h3> <pre><code>1 (1) * 2 (1) * 3 (1) * A (3) *** B (3) *** C (3) *** </code></pre> <h3>Considerations</h3> <p>Here are some things to consider when decided how to solve this problem.</p> <ul> <li>Will you always know what characters need to be tracked ahead of time, or will there be times when you want to track <em>any</em> character? In the latter case, an array won't work for you; you would need to use an advanced data structure like a TreeMap or HashMap.</li> <li>Will you always be counting occurrences of specific <code>char</code>s, as opposed to <code>String</code>s? If you have to modify this to count <code>String</code>s then using the <code>String indexes</code> map trick isn't going to work for you either.</li> <li>Are you learning a specific data structure(s) in your course at the moment? Usually problems such as this are assigned to students to understand how to apply a specific concept. As <strong><em>@kyle</em></strong> suggested, you should try to use the data structure that you are learning about, or have learned about, in your class. <em>Sometimes using structures that you haven't learned about yet can get you in trouble, or a lower grade at least.</em></li> </ul>
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    1. POCounting the occurrence of alphanumeric characters and printing them graphically
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    1. COReally nice what you did to format the output!!
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    2. COI've added another approach... quite an overkill. If I have time I'll create a third one with as less lines of code as possible.
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    3. COWhile it is true that you don't need a `HashMap`, it is a data structure fit for the task (well, actually a MultiSet is, but we don't have that one in Java standard libraries). Your approach works for _most_ of the inputs well, but what if I happen to live in a country that has special characters like "ěščřžýáíé"? What if I forgot some ("ůúňťď")? Also, invoking `indexOf()` on every iteration works, but I am curious about its performance with looooooong input strings. All and all, while this works, I wouldn't say it's a way to go with modern Java approaches existing. This is 1990s calling again
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