<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 {
public:
    #define pr pair <int , int> 
    int findLeastNumOfUniqueInts(vector<int>& arr, int k) {
        unordered_map <int , int > mp;
        vector <pr> vect;
        priority_queue <pr> pq;
        priority_queue <pr , vector <pr>  , greater <pr> > pq1; 
        for(auto &it : arr)
            mp[it]++;
        for(auto &it : mp)
            vect.push_back({it.second , it.first});
        for(auto & it : vect) {
            pq.push(it);
            if(pq.size() > k)
                pq.pop();
        }
        while(!pq.empty()) {
            pq1.push(pq.top());
            pq.pop();
        }
        int removed = 0;
        while(!pq1.empty()) {
            int count = pq1.top().first , val = pq1.top().second;
            if( k >= count) {
                k -= count;
                removed++;
            }
            else
                break;
            pq1.pop();
        }
        return mp.size() - removed;

       
    }
};
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