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    <p>Here's a relatively simple solution that's <code>O(N lg(N))</code> that doesn't rely on the entries being among finitely many integers (in particular, it should work for any ordered data type).</p> <p>We assume the output is to be stored in two arrays; <code>lowleft[i]</code> will at the end contain the number of distinct values <code>x[j]</code> with <code>j &lt; i</code> and <code>x[j] &lt; x[i]</code>, and <code>highright[i]</code> will at the end contain the number of distinct values <code>x[j]</code> with <code>j &gt; i</code> and <code>x[j] &gt; x[i]</code>.</p> <p>Create a balanced tree data structure that maintains in each node, the number of nodes in the subtree rooted at that node. This is fairly standard, but not a part of the Java standard library I think; it's probably easiest to do an AVL tree or so. The type of the values in the nodes should be the type of the values in your array.</p> <p>Now first iterate <em>forward</em> through the array. We start with an empty balanced tree. For every value <code>x[i]</code> we encounter, we enter it into the balanced tree (near the end there are <code>O(N)</code> entries in this tree, so this step takes <code>O(lg(N))</code> time). When searching for the position to enter <code>x[i]</code>, we keep track of the number of values less than <code>x[i]</code> by adding up the sizes of all left subtrees whenever we take the right subtree, and adding what will be the size of the left subtree of <code>x[i]</code>. We enter this number into <code>lowleft[i]</code>.</p> <p>If the value <code>x[i]</code> is already in the tree, we just carry on with the next iteration of this loop. If the value <code>x[i]</code> is not in there, we enter it and rebalance the tree, taking care to update the subtree sizes correctly.</p> <p>Each iteration of this loop takes <code>O(lg(N))</code> steps, for a total of <code>O(N lg(N))</code>. We now start with an empty tree and do the same thing iterating <em>backward</em> through the array, finding the position for every <code>x[i]</code> in the tree, and every time recording the size of all subtrees to the right of the new node as <code>highright[i]</code>. Total complexity therefore <code>O(N lg(N))</code>.</p>
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    1. POCompute smaller and bigger values for an array position
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    1. USErik P.
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    1. COGood one, although it's harder to code.
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      1. USjusthalf
    2. COYes, it would be a good solution but if i understand it correctly there is a problem: for an array 1 5 4 3 6 5 9 10, the first 5 has 6 as a bigger number, but the second 5 doesn't. So for the second five we would get a +1 value that doesn't meet the requirements. I think. Thanks
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      1. USMoises B.
    3. CO@user1812632 For the `highright` entries, you iterate *backward* through the array (in the second pass). So early on you encounter the rightmost 5 and see that there's one number that's greater in your tree (the 9), and later you encounter the leftmost 5 and see that the 6 and 9 are the numbers in your tree that are greater.
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