<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:
        m1=0
        if len(nums)<2:
            return 0
        d={}
        for i in range(0,len(nums),2):
            if nums[i] not in d:
                d[nums[i]]=1
            else:
                d[nums[i]]+=1
        d1={}
        for i in range(1,len(nums),2):
            if nums[i] not in d1:
                d1[nums[i]]=1
            else:
                d1[nums[i]]+=1
        maxi=(max(max(d1.values()),max(d.values())))
        d1=(sorted(d1.items(), key=lambda x:x[1] ,reverse=True))
        d2=(sorted(d.items(), key=lambda x:x[1] ,reverse=True))
        ans=[]
        flag=0
        for i in range(0,len(d1)):
            if i==2:
                break
            key,value=d1[i][0],d1[i][1]
            for j in range(0,len(d2)):
                if j==2:
                    break
                if d2[j][0]!=key:
                    temp=value+d2[j][1]
                    ans.append(temp)
        # print(d1)
        # print(d2)
        if not ans:
            return len(nums)-maxi
        return len(nums)-max(ans)

853