Append K Integers With Minimal Sum
<p>You are given an integer array <code>nums</code> and an integer <code>k</code>. Append <code>k</code> <strong>unique positive</strong> integers that do <strong>not</strong> appear in <code>nums</code> to <code>nums</code> such that the resulting total sum is <strong>minimum</strong>.</p>

<p>Return<em> the sum of the</em> <code>k</code> <em>integers appended to</em> <code>nums</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,4,25,10,25], k = 2
<strong>Output:</strong> 5
<strong>Explanation:</strong> 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.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [5,6], k = 6
<strong>Output:</strong> 25
<strong>Explanation:</strong> 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.
</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>
	<li><code>1 &lt;= k &lt;= 10<sup>8</sup></code></li>
</ul>
Greedy; Array; Math; Sorting
class Solution:
    def minimalKSum(self, nums: List[int], k: int) -> int:
        nums.extend([0, 2 * 10**9])
        nums.sort()
        ans = 0
        for a, b in pairwise(nums):
            m = max(0, min(k, b - a - 1))
            ans += (a + 1 + a + m) * m // 2
            k -= m
        return ans
