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    <p>The transformation you want to do is essentially a circular shift by <code>n</code> (or by <code>m</code>, depending on the direction of the shift).</p> <p>E.g., we have <code>1 2 3 4 5 6 7 a b c</code> (I use letters and numbers to separate two arrays) During this transformation, <code>1</code> will move to the position of <code>4</code>, <code>4</code> will move to <code>7</code>, <code>7</code> to <code>c</code>, <code>c</code> to <code>3</code>, <code>3</code> to <code>6</code>, etc. Eventually, we'll return to the position <code>1</code>, from which started.<br> So, moving one number at the time, we completed it.</p> <p>The only trick is that sometimes we'll return to <code>1</code> before completing whole transformation. Like in the case <code>1 2 a b c d</code>, the positions will be <code>1 -&gt; a -&gt; c -&gt; 1</code>. In this case, we'll need to start from <code>2</code> and repeat operation.</p> <p>You can notice that amount of repetitions we need is a greatest common divisor of <code>n</code> and <code>m</code>.</p> <p>So, the code could look like</p> <pre><code>int repetitions = GCD(n, m); int size = n + m; for (int i = 0; i &lt; repetitions; ++i) { int current_number = a[i]; int j = i; do { j = (j + n) % size; int tmp = current_number; current_number = a[j]; a[j] = tmp; } while (j != i); } </code></pre> <p>Greatest common divisor can be easily computed in <code>O(logn)</code> with <a href="http://en.wikipedia.org/wiki/Greatest_common_divisor#Using_Euclid.27s_algorithm" rel="nofollow">well-known recursive formula</a>.</p> <p><strong>edit</strong><br> It does work, I tried in Java. I only changed data type to string for ease of representation.</p> <pre><code> String[] a = {"1", "2", "3", "4", "5", "6", "a", "b", "c"}; int n = 3; int m = 6; // code from above... System.out.println(Arrays.toString(a)); </code></pre> <p>And Euclid's formula:</p> <pre><code>int GCD(int a, int b) { if (a == 0) { return b; } return GCD(b % a, a); } </code></pre>
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    1. POInterview Question: Replacing two arrays's place in memory
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    1. USNikita Rybak
    1. USNikita Rybak
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    1. COThis is essentially my solution, although I didn't prove that the number of cycles is `GCD(n,m)`. Thanks!
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      1. USElazar Leibovich
    2. CO@Elazar Oh :) It seemed so theoretical, that I couldn't get through it and jumped to the code instead :) +1
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      1. USNikita Rybak
 

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