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    <p>I think that from a big O standpoint, you have it implemented as good as it gets. The overriding cost is the sort, which is O(N log N). One possibility, though, would be to build up a new vector with the duplicate entries rather than use the existing vector with the delete operation removing the non-duplicates. However, this would only be better if the distinct number of duplicates is small relative to the total number of entries. </p> <p>Consider the extreme example. If the original array consisted of 1,000 entries with only one duplicate, then the output would be a vector with just one value. It might be a bit more efficient to create the new vector with one entry rather than deleting the other 999 entries from the original vector. I suspect, however, that in real world testing, the savings of that change could be difficult to measure.</p> <p><strong>Edit</strong> I was just thinking about this in terms of "interview" question. In other words, this is not a terribly useful answer. But it would be possible to solve this in O(N) (linear time) as opposed to O(N Log N). Use storage space instead of CPU. Create two "bit" arrays with them cleared initially. Loop through your vector of integer values. Look up each value in the first bit array. If it is not set, then set the bit (set it to 1). If it is set, then set the corresponding bit in the second array (indicating a duplicate). After all vector entries are processed, scan through the second array and output the integers that are duplicates (indicated by the bits set in the second bit array). The reason for using bit arrays is just for space efficiency. If dealing with 4-byte integers, then the raw space required is <code>(2 * 2^32 / 8 )</code>. But this could actually be turned into a usable algorithm by making it a sparse array. The very pseudo pseudo-code would be something like this:</p> <pre><code>bitarray1[infinite_size]; bitarray2[infinite_size]; clear/zero bitarrays // NOTE - do not need to sort the input foreach value in original vector { if ( bitarray1[value] ) // duplicate bitarray2[value] = 1 bitarray1[value] = 1 } // At this point, bitarray2 contains a 1 for all duplicate values. // Scan it and create the new vector with the answer for i = 0 to maxvalue if ( bitarray2[i] ) print/save/keep i </code></pre>
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    1. POHow to keep only duplicates efficiently?
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    1. USMark Wilkins
    1. USMark Wilkins
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      1. VTUpMod
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