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    <p>I think something like this is what you want.</p> <p>For alternating characters:</p> <pre><code>(?=(.)(?!\1)(.))(?:\1\2){2,} </code></pre> <p><code>\0</code> will be the entire alternating sequence, <code>\1</code> and <code>\2</code> are the two (distinct) alternating characters.</p> <p>For run of N and M characters, possibly separated by other characters (replace <code>N</code> and <code>M</code> with numbers here):</p> <pre><code>(?=(.))\1{N}.*?(?=(?!\1)(.))\2{M} </code></pre> <p><code>\0</code> will be entire match, including infix. <code>\1</code> is the character repeated (at least) <code>N</code> times, <code>\2</code> is the character repeated (at least) <code>M</code> times.</p> <p>Here's a test harness in Java.</p> <pre><code>import java.util.regex.*; public class Regex3 { static String runNrunM(int N, int M) { return "(?=(.))\\1{N}.*?(?=(?!\\1)(.))\\2{M}" .replace("N", String.valueOf(N)) .replace("M", String.valueOf(M)); } static void dumpMatches(String text, String pattern) { Matcher m = Pattern.compile(pattern).matcher(text); System.out.println(text + " &lt;- " + pattern); while (m.find()) { System.out.println(" match"); for (int g = 0; g &lt;= m.groupCount(); g++) { System.out.format(" %d: [%s]%n", g, m.group(g)); } } } public static void main(String[] args) { String[] tests = { "foobababababaf foobaafoobaaaooo", "xxyyyy axxayyyya zzzzzzzzzzzzzz" }; for (String test : tests) { dumpMatches(test, "(?=(.)(?!\\1)(.))(?:\\1\\2){2,}"); } for (String test : tests) { dumpMatches(test, runNrunM(3, 3)); } for (String test : tests) { dumpMatches(test, runNrunM(2, 4)); } } } </code></pre> <p>This produces the following output:</p> <pre><code>foobababababaf foobaafoobaaaooo &lt;- (?=(.)(?!\1)(.))(?:\1\2){2,} match 0: [bababababa] 1: [b] 2: [a] xxyyyy axxayyyya zzzzzzzzzzzzzz &lt;- (?=(.)(?!\1)(.))(?:\1\2){2,} foobababababaf foobaafoobaaaooo &lt;- (?=(.))\1{3}.*?(?=(?!\1)(.))\2{3} match 0: [aaaooo] 1: [a] 2: [o] xxyyyy axxayyyya zzzzzzzzzzzzzz &lt;- (?=(.))\1{3}.*?(?=(?!\1)(.))\2{3} match 0: [yyyy axxayyyya zzz] 1: [y] 2: [z] foobababababaf foobaafoobaaaooo &lt;- (?=(.))\1{2}.*?(?=(?!\1)(.))\2{4} xxyyyy axxayyyya zzzzzzzzzzzzzz &lt;- (?=(.))\1{2}.*?(?=(?!\1)(.))\2{4} match 0: [xxyyyy] 1: [x] 2: [y] match 0: [xxayyyy] 1: [x] 2: [y] </code></pre> <hr> <h3>Explanation</h3> <ul> <li><code>(?=(.)(?!\1)(.))(?:\1\2){2,}</code> has two parts <ul> <li><code>(?=(.)(?!\1)(.))</code> establishes <code>\1</code> and <code>\2</code> using lookahead <ul> <li>Nested negative lookahead ensures that <code>\1</code> != <code>\2</code></li> <li>Using lookahead to capture lets <code>\0</code> have the entire match (instead of just the "tail" end)</li> </ul></li> <li><code>(?:\1\2){2,}</code> captures the <code>\1\2</code> sequence, which must repeat at least twice.</li> </ul></li> <li><code>(?=(.))\1{N}.*?(?=(?!\1)(.))\2{M}</code> has three parts <ul> <li><code>(?=(.))\1{N}</code> captures <code>\1</code> in a lookahead, and then match it <code>N</code> times <ul> <li>Using lookahead to capture means the repetition can be <code>N</code> instead of <code>N-1</code></li> </ul></li> <li><code>.*?</code> allows an infix to separate the two runs, reluctant to keep it as short as possible</li> <li><code>(?=(?!\1)(.))\2{M}</code> <ul> <li>Similar to first part</li> <li>Nested negative lookahead ensures that <code>\1</code> != <code>\2</code></li> </ul></li> </ul></li> </ul> <p>The run regex will match longer runs, e.g. <code>run(2,2)</code> matches <code>"xxxyyy"</code>:</p> <pre><code>xxxyyy &lt;- (?=(.))\1{2}.*?(?=(?!\1)(.))\2{2} match 0: [xxxyy] 1: [x] 2: [y] </code></pre> <p>Also, it does not allow overlapping matches. That is, there is only one <code>run(2,3)</code> in <code>"xx11yyy222"</code>.</p> <pre><code>xx11yyy222 &lt;- (?=(.))\1{2}.*?(?=(?!\1)(.))\2{3} match 0: [xx11yyy] 1: [x] 2: [y] </code></pre>
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