Minimum Difference Subsets!

• Time : $O(sum(A) * n)$
• Space : $O(sum(A))$

int Solution::solve(vector<int> &A) {
    int sum = 0;
    for(auto a : A) sum += a;
    vector<int> dp((sum + 1) / 2 + 1,0);
    dp[0] = 1;
    int res = sum;
    for(auto a : A) {
        for(int j = dp.size() - 1; j >= a; j--) {
            if(dp[j-a]) dp[j] = 1;
            if(dp[j]) res = min(res, abs(sum - 2 * j));
        }
    }
    return res;
}