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    <p>This results in an O(n) database query, and an O(n) greatest time search, for a total complexity of O(2n) (which, of course, is still O(n)):</p> <p>Use a count distinct command in SQL which will 'bin' items for you in minute increments.</p> <p>So you'd run the count query on this table:</p> <pre><code>time 1 2 4 3 3 2 4 1 3 2 </code></pre> <p>And it would return:</p> <pre><code>0 1 1 1 2 3 3 3 4 2 </code></pre> <p>By counting each item.</p> <p>I suspect you can do the same thing with your table, and bin them by the minute, then run an algorithm on that.</p> <pre><code>SELECT customer_name, COUNT(DISTINCT city) as "Distinct Cities" FROM customers GROUP BY customer_name; </code></pre> <p>From this tutorial on count: <a href="http://www.techonthenet.com/sql/count.php" rel="nofollow noreferrer">http://www.techonthenet.com/sql/count.php</a> (near the end).</p> <p>Here is a similar page from MySQL's manual: <a href="http://dev.mysql.com/doc/refman/5.1/en/counting-rows.html" rel="nofollow noreferrer">http://dev.mysql.com/doc/refman/5.1/en/counting-rows.html</a></p> <p>So if you have a table with a timedate in it (to the minute, allowing binning to happen by minutes):</p> <pre><code>datetime (yyyymmddhhmm) 200901121435 200901121538 200901121435 200901121538 200901121435 200901121538 200901121538 200901121435 200901121435 200901121538 200901121435 200901121435 </code></pre> <p>Then the SQL</p> <pre><code>SELECT datetime, COUNT(DISTINCT datetime) as "Date Time" FROM post GROUP BY datetime; </code></pre> <p>should return</p> <pre><code>200901121435 7 200901121538 5 </code></pre> <p>You will still need to post process this, but the hard work of grouping and counting is done, and will only result in just over 500k rows per year (60 minutes, 24 hours, 365 days)</p> <p>The post processing would be:</p> <pre><code>Start at time T = first post time. Set greatestTime = T Sum all counts between T and T+one hour --&gt; currentHourCount and greatestHourCount While records exist past T+one hour Increment T by one minute. While the first element is prior to time T, subtract it while the last element is before time T+ one hour, add it If currentHourCount &gt; greatestHourCount then greatestHourCount = currentHourCount greatestTime = T end while </code></pre> <p>-Adam</p>
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    1. POHow to find the one hour period with the most datapoints?
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    1. USAdam Davis
    1. USAdam Davis
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    1. COThanks. If the "single query" solution above ends up being too hard on the database server I will probably end up using this method, as its a great way to split the work between the database server and the code server.
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    2. CO@recursive: ayup.
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