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    <p>The other answers will work, but <code>groupBy</code> is an expensive operation and inefficient if you have 10k employees and the first two have the same name.</p> <p><strong>1) Short method</strong>: (<code>distinct</code> internally builds a HashSet of all the values)</p> <pre><code>val names = employees.map(_.name) names.distinct != names </code></pre> <p><strong>2) Efficient method</strong>: (only traverses as much of the seq as it needs to before finding dupe)</p> <pre><code>def hasDupes(employees: Seq[Employee]): Boolean = { val names = collection.mutable.Set.empty[String] employees.foreach { e =&gt; if (names(e.name)) return true names += e.name } false } </code></pre> <p>You can write this as a <a href="http://pastebin.com/pdVf213x" rel="nofollow">one-liner</a> using a tail-recursive method, but the fact is that immutable sets are much slower than mutable ones so not much good if we're going for efficiency.</p> <p><strong>3) Off-the-wall method</strong>: (doesn't build a set at all, but starts a quicksort and quits (via a cheap exception) as soon as it finds two names the same. Should be efficient. )</p> <pre><code>class Dupe extends Throwable with util.control.NoStackTrace val dupeOrd = new Ordering[Employee] { def compare(x: Employee, y: Employee) = if (x.name == y.name) throw new Dupe else x.name compare y.name } def hasDupes(employees: Seq[Employee]) = try { employees.sorted(dupeOrd); false } catch { case e: Dupe =&gt; true } </code></pre> <p><strong>Benchmark</strong> of results, with method 0 being om-nom-nom's <code>groupBy</code> one, (time in ms for 1000 runs on 1000 employee Vector, no duplicates):</p> <pre><code>method time Early return if duplicate found 0 (groupBy) 1560 No 1 (distinct) 329 No 2a (mutable.Set) 255 Yes 2b (immutable.Set) 1414 Yes 3 (sorting) 666 Yes //242 if employees already in name order </code></pre>
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