<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:
        n = len(nums)
        odd, even = defaultdict(int), defaultdict(int)
        for i in range(n):
            if i % 2 == 0:
                even[nums[i]] += 1
            else:
                odd[nums[i]] += 1
        topEven, secondEven = (None, 0), (None, 0)
        for num in even:
            if even[num] > topEven[1]:
                topEven, secondEven = (num, even[num]), topEven
            elif even[num] > secondEven[1]:
                secondEven = (num, even[num])
        topOdd, secondOdd = (None, 0), (None, 0)
        for num in odd:
            if odd[num] > topOdd[1]:
                topOdd, secondOdd = (num, odd[num]), topOdd
            elif odd[num] > secondOdd[1]:
                secondOdd = (num, odd[num])
        if topOdd[0] != topEven[0]:
            return n - topOdd[1] - topEven[1]
        else:
            return n - max(secondOdd[1] + topEven[1], secondEven[1] + topOdd[1])
        
## TC & SC: The time complexity is O(n) for counting and finding the top frequencies, and the space complexity is O(k), where n is the length of the array nums and k is the number of unique elements in nums.
905