<p>Given an array of integers&nbsp;<code>arr</code>&nbsp;and an integer <code>k</code>.&nbsp;Find the <em>least number of unique integers</em>&nbsp;after removing <strong>exactly</strong> <code>k</code> elements<b>.</b></p>

<ol>
</ol>

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

<pre>
<strong>Input: </strong>arr = [5,5,4], k = 1
<strong>Output: </strong>1
<strong>Explanation</strong>: Remove the single 4, only 5 is left.
</pre>
<strong class="example">Example 2:</strong>

<pre>
<strong>Input: </strong>arr = [4,3,1,1,3,3,2], k = 3
<strong>Output: </strong>2
<strong>Explanation</strong>: Remove 4, 2 and either one of the two 1s or three 3s. 1 and 3 will be left.</pre>

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

<ul>
	<li><code>1 &lt;= arr.length&nbsp;&lt;= 10^5</code></li>
	<li><code>1 &lt;= arr[i] &lt;= 10^9</code></li>
	<li><code>0 &lt;= k&nbsp;&lt;= arr.length</code></li>
</ul>
class Solution:
    def findLeastNumOfUniqueInts(self, arr: List[int], k: int) -> int:
        map_ = collections.defaultdict(int)
        for n in arr:
            map_[n] += 1
        min_heap = []
        for key, count in map_.items():
            heapq.heappush(min_heap, (count, key))
        while k>0:
            c, key = heapq.heappop(min_heap)
            c -= 1
            k -= 1
            if c > 0:
                heapq.heappush(min_heap, (c, key))
        map_ = collections.defaultdict(int)
        while min_heap:
            c, k = heapq.heappop(min_heap)
            map_[k] = c
        return len(map_)
39200