<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:
        even, odd = [], []
        for i in range(len(nums)):
            if i % 2 == 0:
                even.append(nums[i])
            else:
                odd.append(nums[i])
        
        even_count = Counter(even)
        odd_count = Counter(odd)
        
        evens = sorted(even_count.items(), key=lambda x: -x[1])
        odds = sorted(odd_count.items(), key=lambda x: -x[1])
        
        # Handle cases where either even or odd list is empty
        if not evens:
            return len(nums) - odds[0][1] if odds else 0
        if not odds:
            return len(nums) - evens[0][1]
        
        evennum, evenmax = evens[0]
        oddnum, oddmax = odds[0]
        
        # When the most frequent elements are different
        if evennum != oddnum:
            res = len(even) - evenmax + len(odd) - oddmax
        else:
            # Edge case handling: pick the next most frequent number if available
            even_second_best = evens[1][1] if len(evens) > 1 else 0
            odd_second_best = odds[1][1] if len(odds) > 1 else 0
            
            res = min(
                len(even) - evenmax + len(odd) - odd_second_best,
                len(even) - even_second_best + len(odd) - oddmax
            )
        
        return res
37000