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    <p>The top k problem can be solved in many ways. In your case you add a new parameter, but it really doesn't matter.</p> <p>The first and the easiest one: just sort the array. Time complexity: O(nlogn)</p> <pre><code>public static int[] farthestK(Integer[] a, final int val, int k) { Arrays.sort(a, new java.util.Comparator&lt;Integer&gt;() { @Override public int compare(Integer o1, Integer o2) { return -Math.abs(o1 - val) + Math.abs(o2 - val); } }); int[] c = new int[k]; for (int i = 0; i &lt; k; i++) { c[i] = a[i]; } return c; } </code></pre> <p>The second way: use a heap to save the max k values, Time complexity: O(nlogk)</p> <pre><code>/** * Use a min heap to save the max k values. Time complexity: O(nlogk) */ public static int[] farthestKWithHeap(Integer[] a, final int val, int k) { PriorityQueue&lt;Integer&gt; minHeap = new PriorityQueue&lt;Integer&gt;(4, new java.util.Comparator&lt;Integer&gt;() { @Override public int compare(Integer o1, Integer o2) { return Math.abs(o1 - val) - Math.abs(o2 - val); } }); for (int i : a) { minHeap.add(i); if (minHeap.size() &gt; k) { minHeap.poll(); } } int[] c = new int[k]; for (int i = 0; i &lt; k; i++) { c[i] = minHeap.poll(); } return c; } </code></pre> <p>The third way: divide and conquer, just like quicksort. Partition the array to two part, and find the kth in one of them. Time complexity: O(n + klogk) The code is a little long, so i just provide link here.</p> <p><a href="http://en.wikipedia.org/wiki/Selection_algorithm" rel="nofollow">Selection problem.</a></p>
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    1. COIs there a reason you've changed `int` to `Integer` here?
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      1. USChris Hayes
    2. CO@ChrisHayes Custom comparator could only used to compare Objects, so i have to change int[] to Integer[], you can also do this copy inside the method.
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    3. CO@Override What does this mean? Can you make the code more clear?
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