4 sum II

/*
It is a fact that for any such (i,j,k,l) ith + jth element must be equal to negative of kth +lth element.
we are adding all i and j and put them into mp1 and vice versa for kth and lth elements.
Now we are checking for total numbers of groups possible.
*/
class Solution {
public:
    int fourSumCount(vector<int>& nums1, vector<int>& nums2, vector<int>& nums3, vector<int>& nums4) {
        unordered_map<int,int> mp1,mp2;
        int n=nums1.size();
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                mp1[nums1[i]+nums2[j]]++;
            }
        }
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                mp2[nums3[i]+nums4[j]]++;
            }
        }
        int ans=0;
        for(auto itr:mp1){
            if(mp2[-(itr.first)]) ans+= itr.second*(mp2[-(itr.first)]);
        }
        return ans;

    }
};