<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>
#pragma GCC optimize("O3", "unroll-loops")
class Solution {
public:
    using int2=pair<int, int>;
    struct cmp{
        bool operator()(int2& x, int2& y){
            return x.second>y.second;
        }
    };
    int findLeastNumOfUniqueInts(vector<int>& arr, int k) {
        sort(arr.begin(), arr.end());
        int prev=0;
        vector<int2> freq;
        for(int x: arr){
            if (x!=prev) {
                freq.emplace_back(x, 1);
                prev=x;
            }
            else freq.back().second++;
        }
        priority_queue<int2, vector<int2>, cmp> pq(freq.begin(), freq.end());
        while(k>0 && !pq.empty()){
            auto [x, f]=pq.top();//the min freq[x] on the top
            pq.pop();
            k-=f;
        }
        int sz=pq.size();
    //    cout<<sz<<endl;
        return k==0?sz:sz+1;
    }
};
54400