<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 statistics import mode
from typing import List

class Solution:
    def minimumOperations(self, nums: List[int]) -> int:
        if len(nums) == 1:
            return 0
        
        a = nums[::2]  
        b = nums[1::2]
        
        x = mode(a)

        y = mode(b)
        
        
        if x != y:
            
            return (len(a) - a.count(x)) + (len(b) - b.count(y))
        else:
            second_most_a = None
            max_count_a = 0
            for item in set(a):
                if item != x:
                    count = a.count(item)
                    if count > max_count_a:
                        max_count_a = count
                        second_most_a = item

           
            second_most_b = None
            max_count_b = 0
            for item in set(b):
                if item != y:
                    count = b.count(item)
                    if count > max_count_b:
                        max_count_b = count
                        second_most_b = item

            
            ops1 = (len(a) - a.count(x))  
            if second_most_b is not None:
                ops1 += (len(b) - b.count(second_most_b))  
            else:
                ops1 += len(b)  

            
            ops2 = 0
            if second_most_a is not None:
                ops2 = (len(a) - a.count(second_most_a))  
            else:
                ops2 = len(a)  
            ops2 += (len(b) - b.count(y))  

            
            return min(ops1, ops2)


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