I think this should work.

Let $$$y$$$ be the number of subsets of $$$A$$$ with sum at most $$$K$$$. You can compute this by $$$dp_{i,j} = $$$ number of subsets of $$$0..i$$$ with sum equal to $$$j$$$ in $$$O(NK)$$$.

Then the solution is $$$max(0, 2^N - 2y)$$$.

I'm assuming we can't skip any element from both sets. Then, we know that each person should get at least total of $$$K+1$$$. So, the sum of the array should be $$$ \ge 2 \cdot (K+1) $$$. If not, the answer is 0.

If it's possible, then we need to count the number of subsets of the given array having sum $$$ \le K $$$. Let it be $$$X$$$. Then the answer is $$$2^{n}-2 \cdot X $$$.

We can count the number of subsets having sum $$$ \le K $$$ using $$$O( N \cdot K ) $$$ time and $$$O(K)$$$ space using DP.

$$$ dp[i][j] = dp[i-1][j] + dp[i-1][j-A[i]] $$$, $$$ 1 \le i \le n $$$ , $$$ 0 \le j \le K $$$.

Below is the function:

def solve(l,k):
	n = len(l)
	dp = [0]*(k+1)
	s = sum(l)
	dp[0] = 1
	for i in range(n):
		for j in range(k,l[i]-1,-1):
			dp[j] += dp[j-l[i]]
	return 2**n - 2*sum(dp)

Hey, the constraints seem too tight for an O(NK) solution to pass. Anyone has a faster approach?