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    <p>Here's a (lightly-tested) version of code that does what you asked. You can easily traverse the result in parallel with the inputs to locate insertions and deletions.</p> <pre><code>public class StringDiff { private static int length(String s) { return s == null ? 0 : s.length(); } private static char[] chars(String s) { return s == null ? new char[0] : s.toCharArray(); } private final String left; private final String right; private final char[] lccs; private final String lcs; public StringDiff(String left, String right) { this.left = left; this.right = right; lccs = init(); lcs = new String(lccs); } public String getLcs() { return lcs; } public char[] getLccs() { return lccs.clone(); } private char[] init() { int lLength = length(left); int rLength = length(right); char[] lChars = chars(left); char[] rChars = chars(right); int [][] t = new int [lLength + 1][rLength + 1]; for (int i = lLength - 1; i &gt;= 0; --i) { for (int j = rLength - 1; j &gt;= 0; --j) { if (lChars[i] == rChars[j]) { t[i][j] = t[i + 1][j + 1] + 1; } else { t[i][j] = Math.max(t[i + 1][j], t[i][j + 1]); } } } char[] result = new char[t[0][0]]; int l = 0, r = 0, p = 0; while (l &lt; lLength &amp;&amp; r &lt; rLength) { if (lChars[l] == rChars[r]) { result[p++] = lChars[l++]; r++; } else { if (t[l + 1][r] &gt; t[l][r + 1]) { ++l; } else { ++r; } } } return result; } } </code></pre> <p>According to it, the actual longest subsequence of your original inputs:</p> <pre><code>The quick brown fox jumped over the lazy dog. The quick yellow fox jumped over the well-bred dog. </code></pre> <p>is:</p> <pre><code>The quick ow fox jumped over the l dog. </code></pre> <p>(because "brown" and "yellow" have "ow" in common, etc.)</p> <p>It's relatively straightforward to modify the above to split on whitespace (instead of into char arrays) and substitute String#equals for == to get a version that finds the longest common subsequence of words instead of characters. For your example above that change would produce the obvious result:</p> <pre><code>found 7 words 'The' 'quick' 'fox' 'jumped' 'over' 'the' 'dog.' </code></pre> <p>(Your question implied character comparisons, as you matched the spaces between words.)</p>
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    1. POJava library for free-text diff
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    1. USjoel.neely
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    1. COThat is not what he asked for is it? It should be index to index based and thus not return a match for ow and yellow and brown since they are not on the same index?
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