<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)
        if n == 1:return 0
        freq0 = Counter(nums[0::2])
        freq1 = Counter(nums[1::2])
        l0 = sorted(freq0.keys(),key = lambda x:-freq0[x])
        l1 = sorted(freq1.keys(),key = lambda x:-freq1[x])
        if len(l0) == len(l1) and len(l0) == 1:
            if l0[0] == l1[0]:
                return n // 2
            else:
                return 0

        if len(l0) == 1:
            if l1[0]!=l0[0]:return n - sum(freq0.values()) - freq1[l1[0]]
            else: return min(n - sum(freq0.values()) - freq1[l1[1]], n -  freq1[l1[0]])
        if len(l1) == 1:
            if l1[0]!=l0[0]:return n - sum(freq1.values()) - freq0[l0[0]]
            else: return min(n - sum(freq1.values()) - freq0[l0[1]], n -  freq0[l0[0]])

        if l0[0]!=l1[0]:return n - freq0[l0[0]] -  freq1[l1[0]]
        m0 = freq0[l0[0]]
        m1 = freq1[l1[0]]
        if len(l0)>1:m01 = freq0[l0[1]]
        else:m01 = 0
        if len(l1)>1:m11 = freq1[l1[1]]
        else:m11 = 0
        return min(n-m0-m11,n-m01-m1)
        
        return n // 2
        
        
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