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# Find the K-Sum of an Array LeetCode Solution

## Problem – Find the K-Sum of an Array

You are given an integer array `nums` and a positive integer `k`. You can choose any subsequence of the array and sum all of its elements together.

We define the K-Sum of the array as the `kth` largest subsequence sum that can be obtained (not necessarily distinct).

Return the K-Sum of the array.

subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.

Note that the empty subsequence is considered to have a sum of `0`.

Example 1:

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

Example 2:

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

Constraints:

• `n == nums.length`
• `1 <= n <= 105`
• `-109 <= nums[i] <= 109`
• `1 <= k <= min(2000, 2n)`

### Find the K-Sum of an Array LeetCode Solution in C++

``````class Solution {
public:
long long kSum(vector<int>& nums, int k) {
int n=nums.size();
vector<long long>ans;
priority_queue<pair<long long,long long>>pq;
long long sum=0;
for(int i=0;i<n;i++)
{
if(nums[i]>0)
{
sum+=nums[i];
}
nums[i]=abs(nums[i]);
}
sort(nums.begin(),nums.end());
pq.push({sum-nums,0});
ans.push_back(sum);
while(ans.size()<k)
{
auto [val,index]=pq.top();
pq.pop();
ans.push_back(val);
if(index+1<n)
{
pq.push({val+nums[index]-nums[index+1],index+1});
pq.push({val-nums[index+1],index+1});
}
}
return ans[k-1];
}
};``````

### Find the K-Sum of an Array LeetCode Solution in Java

``````class Solution {
public long kSum(int[] nums, int k) {
long sum = 0;
Queue<Pair<Long, Integer>> pq = new PriorityQueue<>((a, b) -> Long.compare(b.getKey(), a.getKey()));
for (int i = 0; i < nums.length; i++) {
sum = sum + Math.max(0, nums[i]);
}
for (int i = 0; i < nums.length; i++) {
nums[i] = Math.abs(nums[i]);
}
Arrays.sort(nums);
long result = sum;
pq.offer(new Pair<>(sum - nums, 0));
while (--k > 0) {
Pair<Long, Integer> pair = pq.poll();
result = pair.getKey();
int index = pair.getValue();
if (index < nums.length - 1) {
pq.offer(new Pair<>(result + nums[index] - nums[index + 1], index + 1));
pq.offer(new Pair<>(result - nums[index + 1], index + 1));
}
}
return result;
}
}``````

### Find the K-Sum of an Array LeetCode Solution in Python

``````import heapq

class Solution:
def kSum(self, nums: List[int], k: int) -> int:
heap = []

# get maximal subsequence sum
max_subset = sum(n for n in nums if n > 0)
removal_candidates = list(sorted(nums, key=lambda x: abs(x)))

# keep max_heap of next largest subset sum
heap.append((-max_subset, -1))
ans = max_subset

# pop k times
for _ in range(k):
candidate, idx = heapq.heappop(heap)
ans = min(ans, -candidate)

# while we still have items to remove, create new candidate subseq sum
if idx + 1 < len(removal_candidates):
# remove candidate element and keep track of last index
heapq.heappush(heap, (candidate + abs(removal_candidates[idx + 1]), idx + 1))
# if we need to readd elements previously removed
if idx >= 0:
heapq.heappush(heap, (candidate - abs(removal_candidates[idx]) + abs(removal_candidates[idx + 1]), idx + 1))

return ans``````
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