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    <h2>Part 2 of your question</h2> <p>Here is a shorter variant:</p> <pre><code>public String findLongestPrefixSuffix(String s1, String s2) { for( int i = Math.min(s1.length(), s2.length()); ; i--) { if(s2.endsWith(s1.substring(0, i))) { return s1.substring(0, i); } } } </code></pre> <p>I am using <code>Math.min</code> to find the length of the shortest String, as I don't need to and cannot compare more than that.</p> <p><code>someString.substring(x,y)</code> returns you the String you get when reading someString beginning from character <code>x</code> and stopping at character <code>y</code>. I go backwards from the biggest possible substring (<code>s1</code> or <code>s2</code>) to the smallest possible substring, the empty string. This way the first time my condition is true it will be biggest possible substring the fulfills it.</p> <p>If you prefer you can go the other way round, but you have to introduce a variable saving the length of the longest found substring fulfilling the condition so far:</p> <pre><code>public static String findLongestPrefixSuffix(String s1, String s2) { if (s1.equals(s2)) { // this part is optional and will return s1; // speed things up if s1 is equal to s2 } // int max = 0; for (int i = 0; i &lt; Math.min(s1.length(), s2.length()); i++) { if (s2.endsWith(s1.substring(0, i))) { max = i; } } return s1.substring(0, max); } </code></pre> <p>For the record: You could start with <code>i = 1</code> in the latter example for a tiny bit of extra performance. On top of this you can use <code>i</code> to specify how long the suffix has at least to be you want to get. ;) If you writ <code>Math.min(s1.length(), s2.length()) - x</code> you can use <code>x</code> to specify how long the found substring may be at most. Both of these things are possible with the first solution, too, but the min length is a bit more involving. ;)</p> <hr> <h2>Part 1 of your question</h2> <p>In the part above the <code>Collections.reverse</code> the author of the code searches for all positions in <code>s1</code> where the last letter of <code>s2</code> is and saves this position.</p> <p>What follows is essentially what my algorithm does, the difference is, that he doesn't check every substring but only those that end with the last letter of <code>s2</code>.</p> <p>This is some sort of optimization to speed things up. If speed is not that important my naive implementation should suffice. ;)</p>
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    1. COThanks, @TheMorph for such detailed explanation!
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    2. COYou're welcome. ;)
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