Reduce Array Size to The Half
<p>You are given an integer array <code>arr</code>. You can choose a set of integers and remove all the occurrences of these integers in the array.</p>

<p>Return <em>the minimum size of the set so that <strong>at least</strong> half of the integers of the array are removed</em>.</p>

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

<pre>
<strong>Input:</strong> arr = [3,3,3,3,5,5,5,2,2,7]
<strong>Output:</strong> 2
<strong>Explanation:</strong> Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has a size greater than half of the size of the old array.
</pre>

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

<pre>
<strong>Input:</strong> arr = [7,7,7,7,7,7]
<strong>Output:</strong> 1
<strong>Explanation:</strong> The only possible set you can choose is {7}. This will make the new array empty.
</pre>

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

<ul>
	<li><code>2 &lt;= arr.length &lt;= 10<sup>5</sup></code></li>
	<li><code>arr.length</code> is even.</li>
	<li><code>1 &lt;= arr[i] &lt;= 10<sup>5</sup></code></li>
</ul>
Greedy; Array; Hash Table; Sorting; Heap (Priority Queue)
func minSetSize(arr []int) (ans int) {
	mx := slices.Max(arr)
	cnt := make([]int, mx+1)
	for _, x := range arr {
		cnt[x]++
	}
	sort.Ints(cnt)
	for i, m := mx, 0; ; i-- {
		if cnt[i] > 0 {
			m += cnt[i]
			ans++
			if m >= len(arr)/2 {
				return
			}
		}
	}
}