<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:
        n = len(nums)
        if n <= 1:
            return 0
        
        # Count the frequency of even-indexed and odd-indexed numbers
        freq_even = Counter(nums[i] for i in range(0, n, 2))
        freq_odd = Counter(nums[i] for i in range(1, n, 2))
        
        # Find the most common elements for both even and odd indices
        most_common_even = freq_even.most_common(2)
        most_common_odd = freq_odd.most_common(2)
        
        # If the most common elements are different, return n minus the sum of their counts
        if most_common_even[0][0] != most_common_odd[0][0]:
            return n - most_common_even[0][1] - most_common_odd[0][1]
        
        # If the most common elements are the same, we have two options:
        # 1. Replace all even-indexed elements except the most common one
        # 2. Replace all odd-indexed elements except the most common one
        # Choose the option that requires fewer replacements
        else:
            option1 = n - most_common_even[0][1] - (most_common_odd[1][1] if len(most_common_odd) > 1 else 0)
            option2 = n - most_common_odd[0][1] - (most_common_even[1][1] if len(most_common_even) > 1 else 0)
            return min(option1, option2)
837