<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) {

        ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        unordered_map<int, int> factor;
        int value;
        priority_queue<int> min_heap;
        int input = 0;

        for (int n : arr)
            factor[n]++;

        for (auto i : factor) {
            min_heap.push(i.second * -1);
        }

        value = factor.size();

        while (k) {

            if (abs(min_heap.top()) > 1) {

                input = abs(min_heap.top());
                min_heap.pop();
                --input;
                min_heap.push(input * -1);

            } else {
                value--;
                min_heap.pop();
            }
            k--;
        }
        return value;
    }
};
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