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    <p>Given your description, I'm assuming that after the <code>NNDDDDD</code> portion, the first <code>A</code> will actually be a <code>N</code> rather than an <code>A</code>, since otherwise there's no solid boundary between the <code>DDDDD</code> and <code>AAAA</code> portions.</p> <p>So, your string actually looks like <code>NNDDDDDNAAA</code>, and you want to replace the <code>NAAA</code> portion with spaces. Given this, the regex can be rewritten as such: <code>(\\D+\\d+)(\\D.+)</code></p> <p>Positive lookbehind in Java requires a fixed length pattern; You can't use the <code>+</code> or <code>*</code> patterns. You can instead use the curly braces and specify a maximum length. For instance, you can use <code>{1,9}</code> in place of each <code>+</code>, and it will match between 1 and 9 characters: <code>(?&lt;=\\D{1,9}\\d{1,9})(\\D.+)</code></p> <p>The only problem here is you're matching the NAAA sequence as a single match, so using <code>"NNNDDDDNAAA".replaceAll("(?&lt;=\\D{1,9}\\d{1,9})(\\D.+)", " ")</code> will result in replacing the entire <code>NAAA</code> sequence with a single space, rather than multiple spaces.</p> <p>You could take the beginning delimiter of the match, and the string length, and use that to append the correct number of spaces, but I don't see the point. I think you're better off with your original solution; Its simple and easy to follow.</p> <p>If you're looking for a little extra speed, you could compile your Pattern outside the function, and use StringBuilder or StringBuffer to create your output. If you're building a large String out of all these NNDDDDDAAAAA elements, work entirely in StringBuilder until you're done appending.</p> <pre><code>class Test { public static Pattern p = Pattern.compile("(\\D+\\d+)(\\D.+)"); public static StringBuffer replace( String input ) { StringBuffer output = new StringBuffer(); Matcher m = Test.p.matcher(input); if( m.matches() ) output.append( m.group(1) ).append( m.group(2).replaceAll("."," ") ); return output; } public static void main( String[] args ) { String input = args[0]; long startTime; StringBuffer tests = new StringBuffer(); startTime = System.currentTimeMillis(); for( int i = 0; i &lt; 50; i++) { tests.append( "Input -&gt; Output: '" ); tests.append( input ); tests.append( "' -&gt; '" ); tests.append( Test.replace( input ) ); tests.append( "'\n" ); } System.out.println( tests.toString() ); System.out.println( "\n" + (System.currentTimeMillis()-startTime)); } } </code></pre> <p><strong>Update:</strong> I wrote a quick iterative solution, and ran some random data through both. The iterative solution is around 4-5x faster.</p> <pre><code>public static StringBuffer replace( String input ) { StringBuffer output = new StringBuffer(); boolean second = false, third = false; for( int i = 0; i &lt; input.length(); i++ ) { if( !second &amp;&amp; Character.isDigit(input.charAt(i)) ) second = true; if( second &amp;&amp; !third &amp;&amp; Character.isLetter(input.charAt(i)) ) third = true; if( second &amp;&amp; third ) output.append( ' ' ); else output.append( input.charAt(i) ); } return output; } </code></pre>
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