Partition Array Into Two Arrays to Minimize Sum Difference

class Solution {
    int solve(vector<int>& sum1,vector<int>& sum2,int tot){
        int m = sum1.size(),n = sum2.size();
        int minDiff = INT_MAX;
        for(int i=0;i<m;++i){
            int target = tot/2-sum1[i];
            int lb = lower_bound(sum2.begin(),sum2.end(),target)-sum2.begin();
            if(lb==sum2.size())
                minDiff = min(minDiff,abs(tot-2*(sum1[i]+sum2[lb-1])));
            else if(lb==0)
                minDiff = min(minDiff,abs(tot-2*(sum1[i]+sum2[lb])));
            else
                minDiff = min({minDiff,
                               abs(tot-2*(sum1[i]+sum2[lb-1])),
                               abs(tot-2*(sum1[i]+sum2[lb]))});
        }
        return minDiff;
    }
public:
    int minimumDifference(vector<int>& nums) {
        int n=nums.size();
        int tot = 0;
        vector<int> nums1,nums2;
        for(int i=0;i<n/2;++i){
            nums1.push_back(nums[i]);
            nums2.push_back(nums[i+n/2]);
            tot += nums[i]+nums[i+n/2];
        }
        //find all possible sums
        vector<int> sum1[n/2+1],sum2[n/2+1];
        for(int mask=0;mask<(1<<n/2);++mask){
            int tot1 = 0,tot2 = 0,count = 0;
            for(int j=0;j<n/2;++j)
                if(mask&(1<<j)){
                    tot1 += nums1[j];
                    tot2 += nums2[j];
                    count += 1;
                }
            sum1[count].push_back(tot1);
            sum2[count].push_back(tot2);
        }
        for(int i=0;i<=n/2;++i){
            sort(sum1[i].begin(),sum1[i].end());
            sort(sum2[i].begin(),sum2[i].end());
        }
        
        int minDiff = INT_MAX;
        for(int i=0;i<=n/2;++i)
            minDiff = min(minDiff,solve(sum1[i],sum2[n/2-i],tot));
        return minDiff;
    }
};