Least Number of Unique Integers after K Removals
<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>
Greedy; Array; Hash Table; Counting; Sorting
class Solution:
    def findLeastNumOfUniqueInts(self, arr: List[int], k: int) -> int:
        cnt = Counter(arr)
        for i, v in enumerate(sorted(cnt.values())):
            k -= v
            if k < 0:
                return len(cnt) - i
        return 0
