Minimum Operations to Make an Array Continuous

You are given a non-empty integer array nums of length n. In one operation, you may replace any one element with any integer.
An array is continuous when both conditions hold:
All n elements are unique.
The difference between the maximum and minimum elements is exactly n - 1.
Return the minimum number of replacement operations needed to make nums continuous.

Function
minOperations(nums: int[]) → int

Examples
Example 1
nums = [4,2,5,3]
return = 0
The values are already the unique consecutive set {2,3,4,5}.
Example 2
nums = [1,2,3,5,6]
return = 1
Replace 1 with 4 to obtain the continuous set {2,3,4,5,6}.
Example 3
nums = [1,10,100,1000]
return = 3
At most one existing value can be retained inside any interval containing four consecutive integers, so three elements must be replaced.

Reduce it fast. Duplicates are useless, so dedupe and sort. Now any final array is a range [x, x+n-1]. Keep as many distinct existing values inside that range as you can. Answer is n minus the max kept. Use two pointers: for each left index, advance right while sorted[right] <= sorted[left] + n - 1. Track the max window size. The classic pitfall is forgetting to dedupe, which counts repeated values as kept and gives answers that are too low. Another is using the deduped length instead of the original n for the range width. Check example 3: [1,10,100,1000], n=4, every window keeps one value, so the answer is 3. Complexity is O(n log n) from sorting. If your mind goes blank during the live OA, StealthCoder is the hedge that reads the problem and hands you this approach.

Memorize the pattern. If you can't, run StealthCoder. The proctor sees the IDE. They don't see what's behind it.