<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

        even_counter = Counter(nums[::2])
        max_even = heapq.nlargest(2, even_counter, key=lambda x: even_counter[x])
        # print(max_even, even_counter[max_even[0]], even_counter[max_even[1]])
        odd_counter = Counter(nums[1::2])
        max_odd = heapq.nlargest(2, odd_counter, key=lambda x: odd_counter[x])
        if max_even[0] != max_odd[0]:
            return len(nums) - even_counter[max_even[0]] - odd_counter[max_odd[0]]

        if len(max_even) == len(max_odd) == 1:
            return len(nums) // 2

        if len(max_even) == 1:
            return len(nums) - even_counter[max_even[0]] - odd_counter[max_odd[1]]

        if len(max_odd) == 1:
            return len(nums) - even_counter[max_even[1]] - odd_counter[max_odd[0]]

        return len(nums) - max(
            even_counter[max_even[1]] + odd_counter[max_odd[0]],
            even_counter[max_even[0]] + odd_counter[max_odd[1]]
        )
        return len(nums) - max(even_counter.values()) - max(odd_counter.values())

        distinct_nums = set(nums)
        even_counter = Counter(nums[::2])
        odd_counter = Counter(nums[1::2])
        ans = len(nums)
        for n0, n1 in combinations(distinct_nums, 2):
            start_n0 = len(nums) - even_counter[n0] - odd_counter[n1]
            start_n1 = len(nums) - even_counter[n1] - odd_counter[n0]
            ans = min(ans, start_n0, start_n1)

        return ans
853