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copied!<p>In the general case, this method works. It surrounds the earliest substring of any keyword in any given String:</p> <pre><code>public String bracketize() { String chars = ...; // you can put whatever input (such as 'these are some chars *£&$') String keyword = ...; // you can put whatever keyword (such as *£&$^%) String longest = ""; for(int i=0;i<keyword.length()-1;i++) { for(int j=keyword.length(); j>i; j--) { String tempString = keyword.substring(i,j); if(chars.indexOf(tempString) != -1 && tempString.length()>longest.length()) { longest = tempString; } } } if(longest.length() == 0) return chars; // no possible substring of keyword exists in chars, so just return chars String bracketized = chars.substring(0,chars.indexOf(longest))+"("+longest+")"+chars.substring(chars.indexOf(longest)+longest.length()); return bracketized; } </code></pre> <p>The nested <code>for</code> loops check every possible substring of <code>keyword</code> and select the longest one that is contained in the bigger <code>String</code>, <code>chars</code>. For example, if the keyword is Dog, it will check the substrings "Dog", "Do", "D", "og", "o", and "g". It stores this longest possible substring in <code>longest</code> (which is initialized to the empty String). If the length of <code>longest</code> is still 0 after checking every substring, then no such substring of <code>keyword</code> can be found in <code>chars</code>, so the original String, chars, is returned. Otherwise, a new string is returned which is <code>chars</code> with the substring <code>longest</code> surrounded by brackets (parentheses).</p> <p>Hope this helps, let me know if it works.</p>

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