<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) < 2:
            return 0
        even = []
        even_table = defaultdict(int)
        for count, num in enumerate(nums):
            if not count&1:
                even_table[num] += 1
        for num, occurance in even_table.items():
            heapq.heappush(even, [-occurance, num])

        odd_table = defaultdict(int)
        odd = []
        for count, num in enumerate(nums):
            if count&1:
                odd_table[num] += 1
        for num, occurance in odd_table.items():
            heapq.heappush(odd, [-occurance, num])
        even_frequent = heapq.heappop(even)
        odd_frequent = heapq.heappop(odd)
        while (even_frequent[1] == odd_frequent[1]):
            if not even and not odd:
                return int(len(nums)/2)
            elif not even:
                return self.count(nums, even_frequent, heapq.heappop(odd))
            elif not odd:
                return self.count(nums, heapq.heappop(even), odd_frequent)
            else:
                return min([self.count(nums, even_frequent, heapq.heappop(odd)),
                           self.count(nums, heapq.heappop(even), odd_frequent)])
        return self.count(nums, even_frequent, odd_frequent)

    def count(self, nums: List[int], even_frequent: List[int], odd_frequent: List[int]) -> int:
        res = 0
        if not even_frequent or not odd_frequent:
            return float('inf')
        for count, num in enumerate(nums):
            if not count&1 and num != even_frequent[1]:
                res += 1
            elif count&1 and num !=  odd_frequent[1]:
                    res += 1
        return res
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