<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(set(nums))==1:
            return len(nums)//2
        odd = defaultdict(int)
        even = defaultdict(int)

        n=len(nums)
        #num,freq
        
        for i in range(n):
            if i%2==0:
                even[nums[i]]+=1
            else:
                odd[nums[i]]+=1
        
        count = 0
        odd_ls=list(sorted(odd.items(), key = lambda x:-x[1])) + [(0,0)]
        even_ls=list(sorted(even.items(), key = lambda x:-x[1])) + [(0,0)]

        odd_num, odd_freq = odd_ls[0]
        even_num, even_freq = even_ls[0]

        if odd_num == even_num:
            if odd_ls[1][1] + even_freq > even_ls[1][1] + odd_freq:
                odd_num = odd_ls[1][0]
            else:
                even_num = even_ls[1][0]

        for i in range(n):
            if i%2==0:
                if nums[i]!=even_num:
                    count+=1
            else:
                if nums[i]!=odd_num:
                    count+=1
        return count

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