<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:
        """ Goal: Smallest number of numbers to remove
        1) Clarifying questions
        1.1) What if we can't get to n//2?
        2) Inputs: arr --> output: Integer (the count of numbers we will remove)
        3) hashmap<number, frequency>: Example : {3: 4, 5:3, 2: 2, 7:1}
        4) Sort the hashmap by its value (count) from greatest to smallest
        4.1) Remove the most frequent elements untli we get a size of the array that is at least half of the len(origin_arr)
        5) Make sure to keep the global var ( return value )
        """

        # Populate counts hashmap
        counts = {}
        for num in arr:
            counts[num] = counts.get(num, 0) + 1
        

        res, currRemoved = 0, 0# return var 

        # Sort from greatest to least, then we iterate
        n = len(arr)
        for num, count in sorted(counts.items(), key = lambda x: x[1], reverse = True):
            res += 1
            currRemoved += count

            # Odd numbers
            if len(arr) % 2 != 0 and currRemoved > n // 2:
                break

            # Even numbers
            if len(arr) % 2 == 0 and currRemoved >= n // 2:
                break

        return res

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