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    <p>Sorting is an option, at the expense of extra memory. Consider the following algorithm.</p> <pre><code>1. Allocate additional array and copy into - O(n) 2. Sort additional array - O(n lg n) 3. Lop off the top k elements (in this case 5) - O(n), since k could be up to n 4. Iterate over the original array - O(n) 4.a search the top k elements for to see if they contain the current element - O(lg n) </code></pre> <p>So it step 4 is (n * lg n), just like the sort. The entire algorithm is n lg n, and is very simple to code.</p> <p>Here's a quick and dirty example. There may be bugs in it, and obviously null checking and the like come into play.</p> <p>import java.util.Arrays;</p> <pre><code>class ArrayTest { public static void main(String[] args) { int[] arr = {1, 3, 5, 7, 9, 2, 4, 6, 8, 10}; int[] indexes = indexesOfTopElements(arr,3); for(int i = 0; i &lt; indexes.length; i++) { int index = indexes[i]; System.out.println(index + " " + arr[index]); } } static int[] indexesOfTopElements(int[] orig, int nummax) { int[] copy = Arrays.copyOf(orig,orig.length); Arrays.sort(copy); int[] honey = Arrays.copyOfRange(copy,copy.length - nummax, copy.length); int[] result = new int[nummax]; int resultPos = 0; for(int i = 0; i &lt; orig.length; i++) { int onTrial = orig[i]; int index = Arrays.binarySearch(honey,onTrial); if(index &lt; 0) continue; result[resultPos++] = i; } return result; } } </code></pre> <p>There are other things you can do to reduce the overhead of this operation. For example instead of sorting, you could opt to use a queue that just tracks the largest 5. Being <code>int</code>s they values would probably have to be boxed to be added to a collection (unless you rolled your own) which adds to overhead significantly.</p>
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    1. CO@TheNewIdiot - the correct usage is `lg` not `log`. And 4.a is not the same as 5, because it's not a different step - it's what happens during the iteration.
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    2. COthey both mean the same thing...
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    3. CO@Hunter Kind of. `lg` means `log base 2`. When in big `O` notation, they do condense to one another because they are on the same order, that's correct. However, if that change had been reviewed it would have been rejected as "too minor", and the changing of 4.a to 5 would be rejected as "incorrect".
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