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    <p><strong>The <code>O(NK)</code> <a href="http://en.wikipedia.org/wiki/Dynamic_programming" rel="nofollow">dynamic programming</a> solution</strong> is fairly easy:</p> <p>Let <code>A[i]</code> be the best sum of the elements to the left subject to the not-<code>k</code>-consecutive constraint (assuming we're removing the <code>i</code>-th element as well).</p> <p>Then we can calculate <code>A[i]</code> by looking back <code>K</code> elements:</p> <pre><code>A[i] = 0; for j = 1 to k A[i] = max(A[i], A[i-j]) A[i] += input[i] </code></pre> <p>And, at the end, just look through the last <code>k</code> elements from <code>A</code>, adding the elements to the right to each and picking the best one.</p> <p>But this is too slow.</p> <p><strong>Let's do better.</strong></p> <p>So <code>A[i]</code> finds the best from <code>A[i-1]</code>, <code>A[i-2]</code>, ..., <code>A[i-K+1]</code>, <code>A[i-K]</code>.<br> So <code>A[i+1]</code> finds the best from <code>A[i]</code>, <code>A[i-1]</code>, <code>A[i-2]</code>, ..., <code>A[i-K+1]</code>.</p> <p>There's a lot of redundancy there - we already know the best from indices <code>i-1</code> through <code>i-K</code> because of <code>A[i]</code>'s calculation, but then we find the best of all of those except <code>i-K</code> (with <code>i</code>) again in <code>A[i+1]</code>.</p> <p>So we can just store all of them in an ordered data structure and then remove <code>A[i-K]</code> and insert <code>A[i]</code>. My choice - A <a href="http://en.wikipedia.org/wiki/Binary_search_tree" rel="nofollow">binary search tree</a> to find the minimum, along with a <a href="http://en.wikipedia.org/wiki/Circular_buffer" rel="nofollow">circular array</a> of size <code>K+1</code> of tree nodes, so we can easily find the one we need to remove.</p> <p>I swapped the problem around to make it slightly simpler - instead of finding the maximum of remaining elements, I find the minimum of removed elements and then return <code>total sum - removed sum</code>.</p> <p><strong>High-level pseudo-code:</strong></p> <pre><code>for each i in input add (i + the smallest value in the BST) to the BST add the above node to the circular array if it wrapper around, remove the overridden element from the BST // now the remaining nodes in the BST are the last k elements return (the total sum - the smallest value in the BST) </code></pre> <p><strong>Running time:</strong></p> <p><code>O(n log k)</code></p> <p><strong>Java code:</strong></p> <pre><code>int getBestSum(int[] input, int K) { Node[] array = new Node[K+1]; TreeSet&lt;Node&gt; nodes = new TreeSet&lt;Node&gt;(); Node n = new Node(0); nodes.add(n); array[0] = n; int arrPos = 0; int sum = 0; for (int i: input) { sum += i; Node oldNode = nodes.first(); Node newNode = new Node(oldNode.value + i); arrPos = (arrPos + 1) % array.length; if (array[arrPos] != null) nodes.remove(array[arrPos]); array[arrPos] = newNode; nodes.add(newNode); } return sum - nodes.first().value; } </code></pre> <p><code>getBestSum(new int[]{1,2,3,1,6,10}, 2)</code> prints <code>21</code>, as required.</p>
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    1. COWhy was this downvoted?
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    2. COthis should be the selected answer - it correctly mentions the O(nk) solution along with how to optimize it to o(nlog(k)). thanks!
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