Minimum Operations to Make the Array Alternating
<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>
Greedy; Array; Hash Table; Counting
class Solution {
public:
    int minimumOperations(vector<int>& nums) {
        auto f = [&](int i) -> vector<int> {
            int k1 = 0, k2 = 0;
            unordered_map<int, int> cnt;
            for (; i < nums.size(); i += 2) {
                cnt[nums[i]]++;
            }
            for (auto& [k, v] : cnt) {
                if (!k1 || cnt[k1] < v) {
                    k2 = k1;
                    k1 = k;
                } else if (!k2 || cnt[k2] < v) {
                    k2 = k;
                }
            }
            return {k1, !k1 ? 0 : cnt[k1], k2, !k2 ? 0 : cnt[k2]};
        };
        vector<int> a = f(0);
        vector<int> b = f(1);
        int n = nums.size();
        if (a[0] != b[0]) {
            return n - (a[1] + b[1]);
        }
        return n - max(a[1] + b[3], a[3] + b[1]);
    }
};