K-diff Pairs in an Array
<p>Given an array of integers <code>nums</code> and an integer <code>k</code>, return <em>the number of <b>unique</b> k-diff pairs in the array</em>.</p>

<p>A <strong>k-diff</strong> pair is an integer pair <code>(nums[i], nums[j])</code>, where the following are true:</p>

<ul>
	<li><code>0 &lt;= i, j &lt; nums.length</code></li>
	<li><code>i != j</code></li>
	<li><code>|nums[i] - nums[j]| == k</code></li>
</ul>

<p><strong>Notice</strong> that <code>|val|</code> denotes the absolute value of <code>val</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,1,4,1,5], k = 2
<strong>Output:</strong> 2
<strong>Explanation:</strong> There are two 2-diff pairs in the array, (1, 3) and (3, 5).
Although we have two 1s in the input, we should only return the number of <strong>unique</strong> pairs.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,4,5], k = 1
<strong>Output:</strong> 4
<strong>Explanation:</strong> There are four 1-diff pairs in the array, (1, 2), (2, 3), (3, 4) and (4, 5).
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,3,1,5,4], k = 0
<strong>Output:</strong> 1
<strong>Explanation:</strong> There is one 0-diff pair in the array, (1, 1).
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
	<li><code>-10<sup>7</sup> &lt;= nums[i] &lt;= 10<sup>7</sup></code></li>
	<li><code>0 &lt;= k &lt;= 10<sup>7</sup></code></li>
</ul>
Array; Hash Table; Two Pointers; Binary Search; Sorting
func findPairs(nums []int, k int) int {
	ans := make(map[int]struct{})
	vis := make(map[int]struct{})

	for _, x := range nums {
		if _, ok := vis[x-k]; ok {
			ans[x-k] = struct{}{}
		}
		if _, ok := vis[x+k]; ok {
			ans[x] = struct{}{}
		}
		vis[x] = struct{}{}
	}
	return len(ans)
}
