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PO
 

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    <p>Yes, there is a faster way, provided you're not space-constrained. (See <a href="http://en.wikipedia.org/wiki/Space-time_tradeoff" rel="noreferrer">space/time tradeoff</a>.)</p> <p>The algorithm:</p> <p>Just insert all the elements in Set2 into a hashtable (in C# 3.5, that's a <a href="http://msdn.microsoft.com/en-us/library/bb359438.aspx" rel="noreferrer">HashSet&lt;string&gt;</a>), and then go through all the elements of Set1 and check if they're in the hashtable. This method is faster (Θ(m + n) time complexity), but uses O(n) space.</p> <p>Alternatively, just say:</p> <pre><code>bool isSuperset = new HashSet&lt;string&gt;(set2).IsSupersetOf(set1); </code></pre> <hr> <p><strong>Edit 1:</strong></p> <p>For those people concerned about the possibility of duplicates (and hence the misnomer "set"), the idea can easily be extended:</p> <p>Just make a new <code>Dictionary&lt;string, int&gt;</code> representing the count of each word in the super-list (add one to the count each time you see an instance of an existing word, and add the word with a count of 1 if it's not in the dictionary), and then go through the sub-list and decrement the count each time. If every word exists in the dictionary <em>and</em> the count is never zero when you try to decrement it, then the subset is in fact a sub-list; otherwise, you had too many instances of a word (or it didn't exist at all), so it's not a real sub-list.</p> <hr> <p><strong>Edit 2:</strong></p> <p>If the strings are very big and you're concerned about space efficiency, and an algorithm that works with (very) high probability works for you, then try storing a <em>hash</em> of each string instead. It's technically not <em>guaranteed</em> to work, but the probability of it not working is pretty darn low.</p>
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    1. COJust use [`IsSubsetOf`](http://msdn.microsoft.com/en-us/library/bb358446.aspx) :)
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    2. CO@Porges: Edit: I thought you meant that `IsSubsetOf` is a LINQ method, but it wasn't -- is that method really what you meant, or did you mean `IsSupersetOf`? (I think using `IsSubsetOf` on the subset is slower than using `IsSupersetOf` on the superset.)
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    3. COIf you have duplicates, using sets and set theory is not the way to go. "A set is a collection that contains no duplicate elements" and the logic makes this assumption. If you remove the second A1 from Set2, both A1s from Set1 will still be considered as "in" Set2.
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      1. USDavid Ruttka
 

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