<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
        
        # Separate numbers into even and odd indexed groups
        even_counts = Counter(nums[i] for i in range(0, len(nums), 2))
        odd_counts = Counter(nums[i] for i in range(1, len(nums), 2))
        
        # Get the most common elements and their counts
        even_most_common = even_counts.most_common()
        odd_most_common = odd_counts.most_common()
        
        # Add a dummy element with count 0 to handle cases where there are no second most common
        even_most_common.append((None, 0))
        odd_most_common.append((None, 0))
        
        # Calculate the total number of elements in even and odd positions
        even_total = sum(even_counts.values())
        odd_total = sum(odd_counts.values())
        
        # Try to make the most common elements in even and odd positions different
        if even_most_common[0][0] != odd_most_common[0][0]:
            # If the most common elements are different, no conflict
            return even_total - even_most_common[0][1] + odd_total - odd_most_common[0][1]
        else:
            # If they are the same, we need to consider the second most common
            return min(
                even_total - even_most_common[0][1] + odd_total - odd_most_common[1][1],
                even_total - even_most_common[1][1] + odd_total - odd_most_common[0][1]
            )

# Example usage:
# solution = Solution()
# print(solution.minimumOperations([3,1,3,2,4,3]))  # Output: 3
# print(solution.minimumOperations([1,2,2,2,2]))    # Output: 2
36900