<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 Solution:
    def minimumOperations(self, nums: List[int]) -> int:

        evens = [N for i, N in enumerate(nums) if i % 2 == 0]
        odds = [N for i, N in enumerate(nums) if i % 2 == 1]
        # print(f'{evens=}')
        # print(f'{odds=}')
        evenLen = len(evens)
        oddLen = len(odds)
        evenCounts = Counter(evens)
        oddCounts = Counter(odds)
        # print(f'{evenCounts=}')
        # print(f'{oddCounts=}')
        evenReplacements = [
            (evenLen - eCount, eNum)
            for eNum, eCount in evenCounts.items()
        ]
        oddReplacements = [
            (oddLen - oCount, oNum)
            for oNum, oCount in oddCounts.items()
        ]
        evenReplacements.sort()
        evenReplacements = evenReplacements[:2]
        oddReplacements.sort()
        oddReplacements = oddReplacements[:2]
        print(f'{evenReplacements=}')
        print(f'{oddReplacements=}')

        answer = max(0, evenLen + oddLen - 1)   # worst case: replace all but 1
        for eReplace, eNum in evenReplacements:
            print(f'even {eNum}: {eReplace}')
            for oReplace, oNum in oddReplacements:
                print(f'  odd {oNum}: {oReplace}')
                if eNum == oNum:
                    print(f'    Equal')
                    possibles = [
                        eReplace + oddLen,   # change these evens and all odds
                        oReplace + evenLen,  # change all evens and these odds
                    ]
                else:
                    print(f'    Different')
                    possibles = [
                        eReplace + oReplace     # change just this set of evens and odds
                    ]
                print(f'      {possibles=}')
                answer = min(
                    answer,
                    min(possibles)
                )
        return answer

1050