Minimum Number of Operations to Make Array Empty
<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>
Greedy; Array; Hash Table; Counting
function minOperations(nums: number[]): number {
    const count: Map<number, number> = new Map();
    for (const num of nums) {
        count.set(num, (count.get(num) ?? 0) + 1);
    }
    let ans = 0;
    for (const [_, c] of count) {
        if (c < 2) {
            return -1;
        }
        ans += ((c + 2) / 3) | 0;
    }
    return ans;
}
