<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>

class Solution:
    def minOperations(self, nums):
        n = len(nums)
        nums.sort()  # Sort the input nums in non-decreasing order.
        
        if n == 1:
            return 0  # If there's only one element, no operations are needed.
        
        mini = float('inf')  # Initialize the minimum operations to a very large value.
        v = [0] * (n + 1)  # Create a list to store prefix sums.
        mp = {}  # Create a dictionary to store the frequency of each element.

        count = 0  # Initialize a count variable.
        
        # Calculate prefix sums and update the count of repeated elements.
        for i in range(n):
            mp[nums[i]] = mp.get(nums[i], 0) + 1
            if mp[nums[i]] != 1:
                count += 1
            v[i] = count
        v[n] = v[n - 1]  # Set the last element of v to be the same as the second-to-last.

        for i in range(n - 1):
            idx = self.lower_bound(nums, nums[i] + n - 1)  # Find the index of the upper bound.
            x = i

            if idx != n and nums[idx] != nums[i] + n - 1:
                x += 1
            if idx == n:
                x += 1

            x += n - 1 - idx
            x += v[idx] - v[i]  # Calculate the number of operations required.
            mini = min(mini, x)  # Update the minimum operations.

        return mini  # Return the minimum operations required.

    def lower_bound(self, nums, target):
        left, right = 0, len(nums)
        while left < right:
            mid = left + (right - left) // 2
            if nums[mid] < target:
                left = mid + 1
            else:
                right = mid
        return left  # Return the index of the lower bound.
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