<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 binSearch(self, nums, start, end, search_element):
        # for the range, we want to return
        # the index where the search_element will go
        # i.e after all element lesser than it

        while start < end:
            mid = (start + end) // 2

            if nums[mid] < search_element:
                start = mid + 1
            else:
                end = mid
        
        return start



    def minOperations(self, nums: List[int]) -> int:
        
        n = len(nums)

        # remove duplicates and sort
        nums = sorted(set(nums))
        hashmap = {num: idx for idx, num in enumerate(nums)}

        # for each element we run two binary searches
        # we find the number of elements falling in the range

        min_changes = n

        for i in range(len(nums)):
            # if curr number is upper limit, how many elements can we use?
            # i - lw_lim + 1
            target = nums[i] - n + 1

            lw_lim = hashmap[target] if target in hashmap else self.binSearch(nums, 0, i, target)

            min_changes = min(min_changes, n - (i - lw_lim + 1))

            # # if curr number is lower limit, how many elements can we use?
            # # up_lim - i + 1
            # target = nums[i] + n - 1
            # up_lim = hashmap[target] if target in hashmap else self.binSearch(nums, i, len(nums)-1, target)
            # print(up_lim, bisect_right(nums, target))
            # min_changes = min(min_changes, n - (up_lim - i))
        
        return min_changes
901