Find the K-Sum of an Array
<p>You are given an integer array <code>nums</code> and a <strong>positive</strong> integer <code>k</code>. You can choose any <strong>subsequence</strong> of the array and sum all of its elements together.</p>

<p>We define the <strong>K-Sum</strong> of the array as the <code>k<sup>th</sup></code> <strong>largest</strong> subsequence sum that can be obtained (<strong>not</strong> necessarily distinct).</p>

<p>Return <em>the K-Sum of the array</em>.</p>

<p>A <strong>subsequence</strong> is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.</p>

<p><strong>Note</strong> that the empty subsequence is considered to have a sum of <code>0</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [2,4,-2], k = 5
<strong>Output:</strong> 2
<strong>Explanation:</strong> All the possible subsequence sums that we can obtain are the following sorted in decreasing order:
- 6, 4, 4, 2, <u>2</u>, 0, 0, -2.
The 5-Sum of the array is 2.
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,-2,3,4,-10,12], k = 16
<strong>Output:</strong> 10
<strong>Explanation:</strong> The 16-Sum of the array is 10.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == nums.length</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= k &lt;= min(2000, 2<sup>n</sup>)</code></li>
</ul>
Array; Sorting; Heap (Priority Queue)
class Solution:
    def kSum(self, nums: List[int], k: int) -> int:
        mx = 0
        for i, x in enumerate(nums):
            if x > 0:
                mx += x
            else:
                nums[i] = -x
        nums.sort()
        h = [(0, 0)]
        for _ in range(k - 1):
            s, i = heappop(h)
            if i < len(nums):
                heappush(h, (s + nums[i], i + 1))
                if i:
                    heappush(h, (s + nums[i] - nums[i - 1], i + 1))
        return mx - h[0][0]
