<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 T:
  def __init__(self):
    self.count = collections.Counter()
    self.mx = 0
    self.secondMax = 0
    self.maxFreq = 0
    self.secondMaxFreq = 0


class Solution:
  def minimumOperations(self, nums: list[int]) -> int:
    # 0 := odd indices, 1 := even indices
    ts = [T() for _ in range(2)]

    for i, num in enumerate(nums):
      t = ts[i % 2]
      t.count[num] += 1
      freq = t.count[num]
      if freq > t.maxFreq:
        t.maxFreq = freq
        t.mx = num
      elif freq > t.secondMaxFreq:
        t.secondMaxFreq = freq
        t.secondMax = num

    if ts[0].mx == ts[1].mx:
      return len(nums) - max(ts[0].maxFreq + ts[1].secondMaxFreq,
                             ts[1].maxFreq + ts[0].secondMaxFreq)
    return len(nums) - (ts[0].maxFreq + ts[1].maxFreq)
962