Append K Integers With Minimal Sum

You are given an integer array nums and an integer k. Append k unique positive integers that do not appear in nums to nums such that the resulting total sum is minimum.

Return the sum of the k integers appended to nums.

Example 1:

Input: nums = [1,4,25,10,25], k = 2
Output: 5
Explanation: The two unique positive integers that do not appear in nums which we append are 2 and 3.
The resulting sum of nums is 1 + 4 + 25 + 10 + 25 + 2 + 3 = 70, which is the minimum.
The sum of the two integers appended is 2 + 3 = 5, so we return 5.

Example 2:

Input: nums = [5,6], k = 6
Output: 25
Explanation: The six unique positive integers that do not appear in nums which we append are 1, 2, 3, 4, 7, and 8.
The resulting sum of nums is 5 + 6 + 1 + 2 + 3 + 4 + 7 + 8 = 36, which is the minimum.
The sum of the six integers appended is 1 + 2 + 3 + 4 + 7 + 8 = 25, so we return 25.

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 109
• 1 <= k <= 108

SOLUTION:

class Solution:
    def minimalKSum(self, nums: List[int], k: int) -> int:
        nums.sort()
        nums.insert(0, 0)
        nums.append(float('inf'))
        print(nums)
        total = 0
        n = len(nums)
        for i in range(n - 1):
            if k >= nums[i + 1] - nums[i] - 1:
                if nums[i + 1] - nums[i] > 1:
                    total += nums[i + 1] * (nums[i + 1] - 1) // 2
                    total -= nums[i] * (nums[i] + 1) // 2
                    k -= nums[i + 1] - nums[i] - 1
            else:
                end = nums[i] + k
                beg = nums[i]
                total += end * (end + 1) // 2
                total -= beg * (beg + 1) // 2
                break
        return total