Minimum Number of Operations to Make Array Continuous
<p>You are given an integer array <code>nums</code>. In one operation, you can replace <strong>any</strong> element in <code>nums</code> with <strong>any</strong> integer.</p>

<p><code>nums</code> is considered <strong>continuous</strong> if both of the following conditions are fulfilled:</p>

<ul>
	<li>All elements in <code>nums</code> are <strong>unique</strong>.</li>
	<li>The difference between the <strong>maximum</strong> element and the <strong>minimum</strong> element in <code>nums</code> equals <code>nums.length - 1</code>.</li>
</ul>

<p>For example, <code>nums = [4, 2, 5, 3]</code> is <strong>continuous</strong>, but <code>nums = [1, 2, 3, 5, 6]</code> is <strong>not continuous</strong>.</p>

<p>Return <em>the <strong>minimum</strong> number of operations to make </em><code>nums</code><em> </em><strong><em>continuous</em></strong>.</p>

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

<pre>
<strong>Input:</strong> nums = [4,2,5,3]
<strong>Output:</strong> 0
<strong>Explanation:</strong>&nbsp;nums is already continuous.
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,2,3,5,6]
<strong>Output:</strong> 1
<strong>Explanation:</strong>&nbsp;One possible solution is to change the last element to 4.
The resulting array is [1,2,3,5,4], which is continuous.
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,10,100,1000]
<strong>Output:</strong> 3
<strong>Explanation:</strong>&nbsp;One possible solution is to:
- Change the second element to 2.
- Change the third element to 3.
- Change the fourth element to 4.
The resulting array is [1,2,3,4], which is continuous.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>
Array; Hash Table; Binary Search; Sliding Window
func minOperations(nums []int) int {
	sort.Ints(nums)
	n := len(nums)
	m := 1
	for i := 1; i < n; i++ {
		if nums[i] != nums[i-1] {
			nums[m] = nums[i]
			m++
		}
	}
	ans := n
	for i := 0; i < m; i++ {
		j := sort.Search(m, func(k int) bool { return nums[k] > nums[i]+n-1 })
		ans = min(ans, n-(j-i))
	}
	return ans
}