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    <p>Here is a solution that takes O(n log n) time, O(n) auxiliary space, and exactly n MoveToFront operations.</p> <p>Given the input array, A, Make a second array, B, with value/index pairs, like so...</p> <pre><code>7 8 9 11 1 10 -&gt; (7 0) (8 1) (9 2) (11 3) (1 4) (10 5) [this step takes O(n) time and O(n) space] </code></pre> <p>then sort B in descending order of the value of each pair...</p> <pre><code>(7 0) (8 1) (9 2) (11 3) (1 4) (10 5) -&gt; (11 3) (10 5) (9 2) (8 1) (7 0) (1 4) [this step takes O(n log n) time] </code></pre> <p>prepare a binary search tree, T.</p> <p>Now for each element in B do the following...</p> <pre><code>Let I be the index of this element. Let V be the sum of I and the number of elements in T that are greater than I. Do a MoveToFront operation on the Vth element of A. Add I to T. [Both of the operations on T take O(log n) time] </code></pre> <p>Here is this algorithm working on your example array</p> <pre><code>(11 3) I := 3 V := 3 + 0 = 3 MoveToFront(3): 7 8 9 11 1 10 -&gt; 11 7 8 9 1 10 T: () -&gt; (3) (10 5) I := 5 V := 5 + 0 = 5 MoveToFront(5): 11 7 8 9 1 10 -&gt; 10 11 7 8 9 1 T: (3) -&gt; (3 5) (9 2) I := 2 V := 2 + 2 = 4 MoveToFront(4): 10 11 7 8 9 1 -&gt; 9 10 11 7 8 1 T: (3 5) -&gt; (2 3 5) (8 1) I := 1 V := 1 + 3 = 4 MoveToFront(4): 9 10 11 7 8 1 -&gt; 8 9 10 11 7 1 T: (2 3 5) -&gt; (1 2 3 5) (7 0) I := 0 V := 0 + 4 = 4 MoveToFront(4): 8 9 10 11 7 1 -&gt; 7 8 9 10 11 1 T: (1 2 3 5) -&gt; (0 1 2 3 5) (1 4) I := 4 V := 4 + 1 = 5 MoveToFront(5): 7 8 9 10 11 1 -&gt; 1 7 8 9 10 11 T: (0 1 2 3 5) -&gt; (0 1 2 3 4 5) </code></pre> <p>I imagine you can find ways to sort these arrays that use fewer than n MoveToFront/Back operations, but I don't think you can find those in less than O(n log n) time. If those operations are slow, though, then it might be worth using an algorithm that takes more time to plan so you can do fewer of those operations.</p>
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    1. COThat doesn't solve a problem, as author's example returns 2, but your seems to return 5. Am I missing something?
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    2. COThe author asks if we can sort an array more efficiently than n^2 MoveToFront/Back operations. I showed here that we can do it in n MoveToFront/Back operations if we take O(n log n) time to do it, and I postulate that we can not do better than that without taking more than O(n log n) time.
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    3. CONice. This reminds me of http://www.spoj.pl/problems/YODANESS/ which I used a similar tree for. (though i could have used a range tree)
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