<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>& arrr, int k) {
        map<int, int> mpp;
        for (auto it : arrr) {
            mpp[it]++;
        }

        // Extract the elements into a vector of pairs
        vector<pair<int, int>> arr(mpp.begin(), mpp.end());

        // Sort by frequency (second element) in ascending order
        sort(arr.begin(), arr.end(), [](const pair<int, int>& a, const pair<int, int>& b) {
            return a.second < b.second;
        });

        int id = 0;
        int uniqueCount = arr.size(); // Initial number of unique integers

        // Reduce the frequency of elements by removing `k` elements in total
        while (k > 0 && id < arr.size()) {
            if (arr[id].second <= k) {
                k -= arr[id].second;
                uniqueCount--; // One unique element removed
                id++;
            } else {
                break; // If we can't remove all of the current frequency, break
            }
        }

        return uniqueCount;
    }
};

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