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 {
public:
    int findLeastNumOfUniqueInts(vector<int>& arr, int k) {
        unordered_map<int, int> cnt;
        for (int& x : arr) {
            ++cnt[x];
        }
        vector<int> nums;
        for (auto& [_, c] : cnt) {
            nums.push_back(c);
        }
        sort(nums.begin(), nums.end());
        for (int i = 0, m = nums.size(); i < m; ++i) {
            k -= nums[i];
            if (k < 0) {
                return m - i;
            }
        }
        return 0;
    }
};