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    <p>Short answer: no.</p> <p>Longer answer: yes, several mutually incompatible optimal solutions are known. It depends on n, k and what properties of the array you can guarantee.</p> <p>If you know nothing about the array the lower bound of complexity is obviously O(n), because all elements of the source array must be examined to see if they fit in the top 10. If you know anything about the source array which allows elements to be safely skipped you should use that knowledge.</p> <p>Similarly the upper complexity bound is O(n.log(n)) because you could always choose to find the answer by sorting the array (O(n.log(n)) and returning the first 10 items (O(1)).</p> <p>A linear search comparing each item to the tenth-highest found so far and inserting it at the appropriate place in the list of highest-found-so-far items if needed has similar complexity for the average and best-case scenarios and has a worst-case of O(kn) which is significantly better than O(n-squared). For the sizes you estimate I expect this method to perform well.</p> <p>If n were much larger (~10000) and k were increased in the same ratio it would probably be worthwhile implementing the quickselect algorithm. Quickselect performs better the more elements you want. If, however, k were not increased in scale with n you should stick with linear searching. Quickselect &amp; friends modify the original array, so are less suitable if you cannot do this in place because you need a bunch more storage and a lot of copying that the algorithm complexity does not include.</p> <p>If n is huge (~1e20) you would want to find the k largest from each of a number of partitions of the input array and then find the k-largest from the aggregate of those results, so that you were not trying to analyse more data than you can fit in memory at a time, and to allow the operation to be parallelised efficiently.</p>
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