<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 class="example">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 class="example">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 class="example">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>

class Solution:
    def reinitializePermutation(self, n: int) -> int:
        # Initial permutation
        perm = list(range(n))
        original = perm[:]
        count = 0
        
        while True:
            # Create a new array based on the transformation rules
            arr = [0] * n
            for i in range(n):
                if i % 2 == 0:
                    arr[i] = perm[i // 2]
                else:
                    arr[i] = perm[n // 2 + (i - 1) // 2]
            
            # Increment the operation count
            count += 1
            
            # Check if the permutation is back to the original
            if arr == original:
                break
            
            # Assign arr to perm for the next operation
            perm = arr
        
        return count
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