<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 typing import List, Tuple, Optional
from collections import Counter
from heapq import nlargest

def safeGetSums(l: List[Tuple[int, int]]) -> Tuple[Optional[int], List[int]]:
    if len(l) == 0:
        return (None, [0, 0])
    elif len(l) == 1:
        return (l[0][1], [l[0][0], 0])
    else:
        return (l[0][1], [l[0][0], l[1][0]])

class Solution:
    def minimumOperations(self, nums: List[int]) -> int:
        odd = Counter(num for idx, num in enumerate(nums) if idx % 2 == 1) # {0: 1}
        even = Counter(num for idx, num in enumerate(nums) if idx % 2 == 0) # {1: 1}
        largestOdd = nlargest(2, ((v,k) for k,v in odd.items())) # [(1, 0)]
        largestEven = nlargest(2, ((v,k) for k,v in even.items())) # [(1, 1)]
        oddTop, oddSums = safeGetSums(largestOdd) # 0, [1,0]
        evenTop, evenSums = safeGetSums(largestEven) # 1, [1,0]
        if oddTop == evenTop:
            return len(nums) - oddSums[0] - evenSums[0] + min(oddSums[0] - oddSums[1], evenSums[0] - evenSums[1]) # 2 - 1 - 1 + min(1 - 0, 1 - 0)
        else:
            return len(nums) - oddSums[0] - evenSums[0] # 2 - 1 - 1
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