<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>

'''
nums = [3,1,3,2,4,3]

odd = {3:3, 4:1} => maximum occurence 2 number ( a1, a2 )
even = {x1:y1, x2:y2, ...} => maximum occurence 2 number ( b1, b2 )


time complexity: O(N)

'''


class Solution:
    def minimumOperations(self, nums: List[int]) -> int:
        odd = {}
        even = {}
        N = len(nums)
        for i in range(N):
            if i & 1:
                if nums[i] not in odd:
                    odd[nums[i]] = 0
                odd[nums[i]] += 1
            else:
                if nums[i] not in even:
                    even[nums[i]] = 0
                even[nums[i]] += 1
        
        a1 = [0, 0]
        a2 = [0, 0]
        for k,v in odd.items():
            if v > a2[1]:
                a2[0] = k
                a2[1] = v

                if a2[1] > a1[1]:
                    a1, a2 = a2, a1

        b1 = [0, 0]
        b2 = [0, 0]
        for k,v in even.items():
            if v > b2[1]:
                b2[0] = k
                b2[1] = v

                if b2[1] > b1[1]:
                    b1, b2 = b2, b1
        
        ans = N
        if a1[0]==b1[0]:
            ans = N - max(a1[1] + b2[1], a2[1] + b1[1])
        else:
            ans = N - (a1[1] + b1[1])

        return ans

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