<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>

from collections import Counter

class Solution:
    def minimumOperations(self, nums: List[int]) -> int:
        print('new')
        n = len(nums)
        if n == 1:
            return 0
        evens = Counter()
        odds = Counter()
        todds = n // 2
        tevens = n // 2 + (n % 2)
        for i, x in enumerate(nums):
            if i % 2 == 0:
                evens[x] += 1
            else:
                odds[x] += 1
        lst_even = [(-v, k) for k, v in evens.items()]
        lst_odd = [(-v, k) for k, v in odds.items()]
        lst_even.sort()
        lst_odd.sort()
        if lst_even[0][1] != lst_odd[0][1]:
            return tevens + lst_even[0][0] + todds + lst_odd[0][0]
        if len(lst_even) == len(lst_odd) == 1:
            return min(todds, tevens)
        elif len(lst_even) == 1:
            return min(todds + lst_odd[1][0], todds + lst_odd[0][0] + tevens)
        elif len(lst_odd) == 1:
            return min(tevens + lst_even[1][0], tevens + lst_even[0][0] + todds)
        return tevens + todds + min(lst_even[0][0] + lst_odd[1][0], lst_even[1][0] + lst_odd[0][0])

1054