Least Number of Unique Integers after K Removals
<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>
Greedy; Array; Hash Table; Counting; Sorting
func findLeastNumOfUniqueInts(arr []int, k int) int {
	cnt := map[int]int{}
	for _, x := range arr {
		cnt[x]++
	}
	nums := make([]int, 0, len(cnt))
	for _, v := range cnt {
		nums = append(nums, v)
	}
	sort.Ints(nums)
	for i, v := range nums {
		k -= v
		if k < 0 {
			return len(nums) - i
		}
	}
	return 0
}