<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>

class Solution:
    def minSetSize(self, arr: List[int]) -> int:
        hashmap = {}
        half_array_len = len(arr) // 2

        for num in arr:
            if num not in hashmap:
                hashmap[num] = 1
            else:
                hashmap[num] += 1
        
        sorted_by_frequency = sorted(hashmap.items(), key=lambda x: (x[1], x[0]), reverse=True)
        
        amount_in_set = 0
        remaining = len(arr)

        for key, value in sorted_by_frequency:
            amount_in_set += 1
            remaining -= value
            if remaining <= half_array_len:
                return amount_in_set

43403