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copied!<p>Jon Skeet's / Timwi's solutions are the way to go.</p> <p>For fun, here's a LINQ query that does the job (<em>very</em> inefficiently):</p> <pre><code>var sequences = input.Distinct() .GroupBy(num => Enumerable.Range(num, int.MaxValue - num + 1) .TakeWhile(input.Contains) .Last()) //use the last member of the consecutive sequence as the key .Where(seq => seq.Count() >= 3) .Select(seq => seq.OrderBy(num => num)); // not necessary unless ordering is desirable inside each sequence. </code></pre> <p>The query's performance can be improved slightly by loading the input into a <code>HashSet</code> (to improve <code>Contains</code>), but that will still not produce a solution that is anywhere close to efficient.</p> <p>The only bug I am aware of is the possibility of an arithmetic overflow if the sequence contains negative numbers <s>of large magnitude</s> (we cannot represent the <code>count</code> parameter for <code>Range</code>). This would be easy to fix with a custom <code>static IEnumerable<int> To(this int start, int end)</code> extension-method. If anyone can think of any other simple technique of dodging the overflow, please let me know.</p> <p>EDIT: Here's a slightly more verbose (but equally inefficient) variant without the overflow issue.</p> <pre><code>var sequences = input.GroupBy(num => input.Where(candidate => candidate >= num) .OrderBy(candidate => candidate) .TakeWhile((candidate, index) => candidate == num + index) .Last()) .Where(seq => seq.Count() >= 3) .Select(seq => seq.OrderBy(num => num)); </code></pre>

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