<p>You are given a <strong>0-indexed</strong> array <code>nums</code> consisting of positive integers.</p>

<p>There are two types of operations that you can apply on the array <strong>any</strong> number of times:</p>

<ul>
	<li>Choose <strong>two</strong> elements with <strong>equal</strong> values and <strong>delete</strong> them from the array.</li>
	<li>Choose <strong>three</strong> elements with <strong>equal</strong> values and <strong>delete</strong> them from the array.</li>
</ul>

<p>Return <em>the <strong>minimum</strong> number of operations required to make the array empty, or </em><code>-1</code><em> if it is not possible</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,3,3,2,2,4,2,3,4]
<strong>Output:</strong> 4
<strong>Explanation:</strong> We can apply the following operations to make the array empty:
- Apply the first operation on the elements at indices 0 and 3. The resulting array is nums = [3,3,2,4,2,3,4].
- Apply the first operation on the elements at indices 2 and 4. The resulting array is nums = [3,3,4,3,4].
- Apply the second operation on the elements at indices 0, 1, and 3. The resulting array is nums = [4,4].
- Apply the first operation on the elements at indices 0 and 1. The resulting array is nums = [].
It can be shown that we cannot make the array empty in less than 4 operations.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,1,2,2,3,3]
<strong>Output:</strong> -1
<strong>Explanation:</strong> It is impossible to empty the array.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Note:</strong> This question is the same as <a href="https://leetcode.com/problems/minimum-rounds-to-complete-all-tasks/description/" target="_blank">2244: Minimum Rounds to Complete All Tasks.</a></p>

class Solution {
public:
    int minOperations(vector<int>& nums) {
        ios::sync_with_stdio(false);
        cin.tie(nullptr);
        cout.tie(nullptr);
        sort(nums.begin(), nums.end());
        int n = nums.size();
        int cnt = 1;
        int ans = 0;
        for(int i = 1; i<n; ++i){
            if(nums[i] == nums[i-1]) { cnt++; continue;}
            if(cnt < 2 ) return -1;
            int threeElement = cnt / 3;
            int numLeft = cnt - threeElement * 3;
            if(numLeft == 0) ans += threeElement;
            else if(numLeft == 1){
                threeElement--;
                ans += threeElement + 2;
            } else ans += threeElement + 1;
            cnt = 1;
        }
        if(cnt < 2) return -1;
        int threeElement = cnt / 3;
            int numLeft = cnt - threeElement * 3;
            if(numLeft == 0) ans += threeElement;
            else if(numLeft == 1){
                threeElement--;
                ans += threeElement + 2;
            } else ans += threeElement + 1;
        return ans;
    }
};
91600