<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:
    int findLeastNumOfUniqueInts(vector<int>& arr, int k) {
        sort(arr.begin(), arr.end());
        map<int,int> mp;
        int prev=arr[0], cnt=0, n=arr.size(), distinct=0;
        for(int i=0;i<n;i++)
        {
            if(arr[i] == prev)
                cnt++;
            else
            {
                distinct++;
                mp[cnt]++;
                cnt=1;
                prev=arr[i];
            }
        }
        mp[cnt]++;
        distinct++;
        int removed=0;
        for(auto it:mp)
        {
            int freq = it.first;
            int ele = it.second;
            int totalEle = freq*ele;
            if(k >= totalEle)
            {
                k-=totalEle;
                removed+=ele;
            }
            else
            {
                removed+=(k/freq);
                k=0;
            }
            if(k == 0)
                break;
        }
        return distinct-removed;
    }
};
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