<p>You are given a <strong>0-indexed</strong> array <code>nums</code> consisting of <code>n</code> positive integers.</p>

<p>The array <code>nums</code> is called <strong>alternating</strong> if:</p>

<ul>
	<li><code>nums[i - 2] == nums[i]</code>, where <code>2 &lt;= i &lt;= n - 1</code>.</li>
	<li><code>nums[i - 1] != nums[i]</code>, where <code>1 &lt;= i &lt;= n - 1</code>.</li>
</ul>

<p>In one <strong>operation</strong>, you can choose an index <code>i</code> and <strong>change</strong> <code>nums[i]</code> into <strong>any</strong> positive integer.</p>

<p>Return <em>the <strong>minimum number of operations</strong> required to make the array alternating</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,1,3,2,4,3]
<strong>Output:</strong> 3
<strong>Explanation:</strong>
One way to make the array alternating is by converting it to [3,1,3,<u><strong>1</strong></u>,<u><strong>3</strong></u>,<u><strong>1</strong></u>].
The number of operations required in this case is 3.
It can be proven that it is not possible to make the array alternating in less than 3 operations. 
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,2,2,2]
<strong>Output:</strong> 2
<strong>Explanation:</strong>
One way to make the array alternating is by converting it to [1,2,<u><strong>1</strong></u>,2,<u><strong>1</strong></u>].
The number of operations required in this case is 2.
Note that the array cannot be converted to [<u><strong>2</strong></u>,2,2,2,2] because in this case nums[0] == nums[1] which violates the conditions of an alternating array.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution:
    def minimumOperations(self, nums: List[int]) -> int:
        n = len(nums)
        if n == 1:
            return 0

        freqs = [Counter(), Counter()]
        counts = [math.ceil(n / 2), n // 2]
        min_ops = n
        
        for i, a in enumerate(nums):
            freqs[i % 2][a] += 1
        
        sorted_nums = [
            sorted([a for a in freqs[0].keys()], key=lambda x: freqs[0][x]),
            sorted([a for a in freqs[1].keys()], key=lambda x: freqs[1][x])
        ]
        
        for a in nums:
            new_ops = []
            for i, j in [[0, 1], [1, 0]]:
                ops = counts[i] - freqs[i][a]
                if sorted_nums[j][-1] == a:
                    ops += counts[j] if len(sorted_nums[j]) == 1 else counts[j] - freqs[j][sorted_nums[j][-2]]
                else:
                    ops += counts[j] - freqs[j][sorted_nums[j][-1]]
                new_ops.append(ops)
            min_ops = min(min_ops, min(new_ops))
        
        return min_ops



        
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