<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:
        
        if len(nums) == 1:
            return 0

        odd_count = {}
        for i in range(len(nums)):
            if i%2 == 1:
                if nums[i] in odd_count:
                    odd_count[nums[i]] += 1
                else:
                    odd_count[nums[i]] = 1
        
        even_count = {}
        for i in range(len(nums)):
            if i%2 == 0:
                if nums[i] in even_count:
                    even_count[nums[i]] += 1
                else:
                    even_count[nums[i]] = 1
        
        odd_count = sorted([(v, k) for k, v in odd_count.items()])
        even_count = sorted([(v, k) for k, v in even_count.items()])
        
        if len(nums)%2 == 0:
            num_even = len(nums)//2
            num_odd = len(nums)//2
        else:
            num_even = len(nums)//2 + 1
            num_odd = len(nums)//2
            
        if odd_count[-1][1] != even_count[-1][1]:
            ops = len(nums) - odd_count[-1][0] - even_count[-1][0]
        else:
            if len(even_count) == 1 and len(odd_count) == 1:
                c1 = odd_count[-1][0]
                c2 = even_count[-1][0]
            elif len(even_count) == 1:
                c1 = even_count[-1][0] + num_odd - odd_count[-1][0]
                c2 = num_odd - odd_count[-2][0]
            elif len(odd_count) == 1:
                c1 = odd_count[-1][0] + num_even - even_count[-1][0]
                c2 = num_even - even_count[-2][0]
            else:
                c1 = len(nums) - odd_count[-1][0] - even_count[-2][0]
                c2 = len(nums) - odd_count[-2][0] - even_count[-1][0]
            
            ops = min(c1, c2)
        return ops

            


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