lec10_46.cpp

class Solution {
    public:
    int minimumDifference(vector<int>& nums) {
        int n = nums.size(),  sum = 0;
        sum = accumulate(nums.begin(), nums.end(),0);  // To find the total sum of the array 
        
        int N = n/2; // Divide it into two equals parts as length is even
        vector<vector<int>> left(N+1), right(N+1); // left array and right array
        
        //All possible sum in left and right part (Generating and storing using BIT-Masking)
        for(int mask = 0; mask<(1<<N); ++mask){  // (1<<n) means 2^n i.e we'll iterate through 0 to 2^n
            int idx = 0, l_sum = 0, r_sum = 0;
            for(int i=0; i<N; ++i){
                if(mask&(1<<i)){  // To check if the bit is set or not 
                    idx++;
                    l_sum += nums[i];
                    r_sum += nums[i+N];
                }
            }
            left[idx].push_back(l_sum);
            right[idx].push_back(r_sum);   // storing
        }
    
        for(int idx=0; idx<=N; ++idx){
            sort(right[idx].begin(), right[idx].end());   // as we'll perform binary search on right so we have to sort it first
        }
    
     int res = min(abs(sum-2*left[N][0]), abs(sum-2*right[N][0]));  // To get the minimum answer from the max sum from both array
            //iterating over left part
            for(int idx=1; idx<N; ++idx){ // iterate from 1 so we dont have to include 0 and check for all value except last as we have alr considered it
                for(auto &a : left[idx]){ // check the sum at each number position
                    int b = (sum - 2*a)/2; // find the value to be minimized 
                    int rightidx = N-idx; // find the number value in right array
                    auto &v = right[rightidx]; // store the vector in v at right number position
                    auto itr = lower_bound(v.begin(), v.end(),b); //binary search over right part
    
                    if(itr!=v.end()) res = min(res, abs(sum-2*(a+(*itr)))); // if found in vector then only update otherwise continue
    
                    }                
                }
            return res;
    
        }
    };