<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>

from collections import Counter
from typing import List

class Solution:
    def minimumOperations(self, nums: List[int]) -> int:
        if len(nums) == 1:
            return 0
        
        freq_even = Counter(nums[i] for i in range(0, len(nums), 2))
        freq_odd = Counter(nums[i] for i in range(1, len(nums), 2))
        
        sorted_even = sorted(freq_even.items(), key=lambda x: -x[1])
        sorted_odd = sorted(freq_odd.items(), key=lambda x: -x[1])
        
        if len(sorted_even) < 2:
            sorted_even.append((0, 0))
        if len(sorted_odd) < 2:
            sorted_odd.append((0, 0))
        
        most_frequent_even, second_most_frequent_even = sorted_even[0], sorted_even[1]
        most_frequent_odd, second_most_frequent_odd = sorted_odd[0], sorted_odd[1]
        
        total_length = len(nums)
        
        if most_frequent_even[0] != most_frequent_odd[0]:
            return total_length - (most_frequent_even[1] + most_frequent_odd[1])
        
        option1 = total_length - (most_frequent_even[1] + second_most_frequent_odd[1])
        option2 = total_length - (second_most_frequent_even[1] + most_frequent_odd[1])
        return min(option1, option2)
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