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leetcode/problem/minimum-number-of-operations-to-reinitialize-a-permutation.html

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+<p>You are given an <strong>even</strong> integer <code>n</code>​​​​​​. You initially have a permutation <code>perm</code> of size <code>n</code>​​ where <code>perm[i] == i</code>​ <strong>(0-indexed)</strong>​​​​.</p>
+
+<p>In one operation, you will create a new array <code>arr</code>, and for each <code>i</code>:</p>
+
+<ul>
+	<li>If <code>i % 2 == 0</code>, then <code>arr[i] = perm[i / 2]</code>.</li>
+	<li>If <code>i % 2 == 1</code>, then <code>arr[i] = perm[n / 2 + (i - 1) / 2]</code>.</li>
+</ul>
+
+<p>You will then assign <code>arr</code>​​​​ to <code>perm</code>.</p>
+
+<p>Return <em>the minimum <strong>non-zero</strong> number of operations you need to perform on </em><code>perm</code><em> to return the permutation to its initial value.</em></p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 2
+<strong>Output:</strong> 1
+<strong>Explanation:</strong> perm = [0,1] initially.
+After the 1<sup>st</sup> operation, perm = [0,1]
+So it takes only 1 operation.
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 4
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> perm = [0,1,2,3] initially.
+After the 1<sup>st</sup> operation, perm = [0,2,1,3]
+After the 2<sup>nd</sup> operation, perm = [0,1,2,3]
+So it takes only 2 operations.
+</pre>
+
+<p><strong>Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 6
+<strong>Output:</strong> 4
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>2 &lt;= n &lt;= 1000</code></li>
+	<li><code>n</code>​​​​​​ is even.</li>
+</ul>