花花酱 LeetCode 2009. Minimum Number of Operations to Make Array Continuous

You are given an integer array nums. In one operation, you can replace any element in nums with any integer.

nums is considered continuous if both of the following conditions are fulfilled:

• All elements in nums are unique.
• The difference between the maximum element and the minimum element in nums equals nums.length - 1.

For example, nums = [4, 2, 5, 3] is continuous, but nums = [1, 2, 3, 5, 6] is not continuous.

Return the minimum number of operations to make numscontinuous.

Example 1:

Input: nums = [4,2,5,3]
Output: 0
Explanation: nums is already continuous.

Example 2:

Input: nums = [1,2,3,5,6]
Output: 1
Explanation: One possible solution is to change the last element to 4.
The resulting array is [1,2,3,5,4], which is continuous.

Example 3:

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

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 109

Solution: Sliding Window

Remove duplicates and sort the numbers.
Try using nums[i] as the min number of the final array.
window [i, j), max – min < n, then change the rest of array to fit into or append after the window, which takes n – (j – i) steps.
e.g. input = [10, 3, 1, 4, 5, 6, 6, 6, 11, 15] => sorted + unique => [1, 3, 4, 5, 6, 10, 11, 15]
n = 10, window = [3, 4, 5, 6, 10, 11], max = 11, min = 3, max – min = 8 < 10
Final array = [3, 4, 5, 6, 1->7, 6₂->8, 6₃->9, 10, 11, 15->12]
Time complexity: O(n)
Space complexity: O(1)