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    <p>For a space efficient method and all values are distinct you can do it in space <code>O( k )</code> and time <code>O( k*log(N)*N )</code>. It's space efficient and there's no data moving and all operations are elementary (adding subtracting).</p> <ol> <li>set <code>U = N; L=0</code></li> <li>First partition the number space in <code>k</code> regions. Like this: <ul> <li><code>0-&gt;(1/k)*(U-L) + L</code>, <code>0-&gt;(2/k)*(U-L) + L</code>, <code>0-&gt;(3/k)*(U-L) + L</code> ... <code>0-&gt;(U-L) + L</code></li> </ul></li> <li>Find how many numbers (<code>count{i}</code>) are in each region. (<code>N*k</code> steps)</li> <li>Find the first region (<code>h</code>) that isn't full. That means <code>count{h} &lt; upper_limit{h}</code>. (<code>k</code> steps)</li> <li>if <code>h - count{h-1} = 1</code> you've got your answer</li> <li>set <code>U = count{h}; L = count{h-1}</code></li> <li>goto 2</li> </ol> <p>this can be improved using hashing (thanks for Nic this idea).</p> <ol> <li>same</li> <li>First partition the number space in <code>k</code> regions. Like this: <ul> <li>`L + (i/k)->L + (i+1/k)*(U-L)'</li> </ul></li> <li><code>inc count{j}</code> using <code>j = (number - L)/k</code> <code>(if L &lt; number &lt; U)</code></li> <li>find first region (<code>h</code>) that doesn't have k elements in it</li> <li>if <code>count{h} = 1</code> h is your answer</li> <li>set <code>U = maximum value in region h</code> <code>L = minimum value in region h</code></li> </ol> <p>This will run in <code>O(log(N)*N)</code>.</p>
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    1. COI really like this answer. It was a bit hard to read, but it's very similar to what I had in my head when I read the question.
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    2. COalso at some point it would be smart to switch to that bitmap solution by Stephen C. probably when `U-L < k`
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