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  1. PO
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    <p>Take a look at the following two methods, they do exactly what you asked for. I wrote them to be generic, it doesn't matter how long your lists are or how many keys exist in the map, the combinations generated are correct.</p> <p>The code below is <em>iterative</em>, based on the algorithm of Python's <a href="http://docs.python.org/library/itertools.html#itertools.product" rel="nofollow"><code>itertools.product()</code></a> function for calculating the Cartesian product of a list of lists.</p> <pre><code>public String[][] allUniqueCombinations() { List&lt;String&gt; labels = new ArrayList&lt;String&gt;(); List&lt;List&lt;String&gt;&gt; lists = new ArrayList&lt;List&lt;String&gt;&gt;(); for (Map.Entry&lt;String, Vector&lt;String&gt;&gt; entry : dataStructure.entrySet()) { labels.add(entry.getKey()); lists.add(entry.getValue()); } List&lt;List&lt;String&gt;&gt; combinations = product(lists); int m = combinations.size() + 1; int n = labels.size(); String[][] answer = new String[m][n]; for (int i = 0; i &lt; n; i++) answer[0][i] = labels.get(i); for (int i = 1; i &lt; m; i++) for (int j = 0; j &lt; n; j++) answer[i][j] = combinations.get(i-1).get(j); return answer; } private List&lt;List&lt;String&gt;&gt; product(List&lt;List&lt;String&gt;&gt; lists) { List&lt;List&lt;String&gt;&gt; result = new ArrayList&lt;List&lt;String&gt;&gt;(); result.add(new ArrayList&lt;String&gt;()); for (List&lt;String&gt; e : lists) { List&lt;List&lt;String&gt;&gt; tmp1 = new ArrayList&lt;List&lt;String&gt;&gt;(); for (List&lt;String&gt; x : result) { for (String y : e) { List&lt;String&gt; tmp2 = new ArrayList&lt;String&gt;(x); tmp2.add(y); tmp1.add(tmp2); } } result = tmp1; } return result; } </code></pre> <p>I tested them with the example in the question:</p> <pre><code>LinkedHashMap&lt;String, Vector&lt;String&gt;&gt; sample = new LinkedHashMap&lt;String, Vector&lt;String&gt;&gt;(); Vector&lt;String&gt; v1 = new Vector&lt;String&gt;(); v1.add("1"); v1.add("2"); v1.add("3"); Vector&lt;String&gt; v2 = new Vector&lt;String&gt;(); v2.add("3"); v2.add("2"); Vector&lt;String&gt; v3 = new Vector&lt;String&gt;(); v3.add("5"); v3.add("6"); v3.add("7"); sample.put("foo", v1); sample.put("bar", v2); sample.put("baz", v3); Foo foo = new Foo(sample); String[][] ans = foo.allUniqueCombinations(); for (String[] row : ans) System.out.println(Arrays.toString(row)); </code></pre> <p>The answer that gets printed is the expected (although the combinations appear in a different order):</p> <pre><code>[foo, bar, baz] [1, 3, 5] [1, 3, 6] [1, 3, 7] [1, 2, 5] [1, 2, 6] [1, 2, 7] [2, 3, 5] [2, 3, 6] [2, 3, 7] [2, 2, 5] [2, 2, 6] [2, 2, 7] [3, 3, 5] [3, 3, 6] [3, 3, 7] [3, 2, 5] [3, 2, 6] [3, 2, 7] </code></pre>
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    1. POJava : Cartesian Product of a List of Lists
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    1. USÓscar López
    1. USÓscar López
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    1. COOscar, it was hard to decide whether to accept yours or Bart's answer, I had to go with Bart's since he answered first, but yours is very good as well, thank you!
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      1. USChris Drappier
 

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