Array - Minimum Operations to Make Array Continuous

Find minimum operations to make array continuous (all distinct, max-min = n-1).

Input: nums=[4,2,5,3] → Output: 0

Input: nums=[1,2,3,5,6] → Output: 1

Input: nums=[1,10,100,1000] → Output: 3

Sort and deduplicate. Use sliding window to find max elements that can be kept in window [x, x+n-1].

Sort and remove duplicates
For each unique value as window start x: window = [x, x+n-1]
Binary search or two-pointer to count elements in window
Operations = n - count (replace elements outside window)
Return minimum operations

public int minOperations(int[] nums) { int n = nums.length; Arrays.sort(nums); int[] unique = Arrays.stream(nums).distinct().toArray(); int m = unique.length, max = 0; int j = 0; for (int i = 0; i < m; i++) { while (j < m && unique[j] < unique[i] + n) j++; max = Math.max(max, j - i); } return n - max; }

Time Complexity: O(n log n)
Space Complexity: O(n)