Leetcode 4Sum II problem solution

from itertools import product
class Solution:
    def fourSumCount(self, A: List[int], B: List[int], C: List[int], D: List[int]) -> int:
        def merge(x, y):
            rest = {}
            for i in product(x, y):
                m = sum(i)
                if m in rest: rest[m] += 1
                else: rest[m] = 1
            return rest
        r1 = merge(A, B)
        r2 = merge(C, D)
        rest = 0
        for i1, i2 in product(r1.keys(), r2.keys()):
            m = i1 + i2
            if m == 0:
                rest += r1[i1] * r2[i2]
        return rest

class Solution {
public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {

    int count=0;
 
    
    HashMap<Integer,Integer> map=new HashMap<>();
    
    for(int i=0;i<A.length;i++){
        for(int j=0;j<B.length;j++){
            
            int sum=A[i]+B[j];
            
            map.put(sum, map.getOrDefault(sum,0)+1);
        }
    }
    
    for(int i=0;i<C.length;i++){
        for(int j=0;j<D.length;j++){
            
            int sum=C[i]+D[j];
            
            if(map.containsKey(-1*sum)){
                
                count+=map.get(-1*sum);
            }
        }
    }
    
    return count;
}
}

class Solution {
public:
    int fourSumCount(vector<int>& nums1, vector<int>& nums2, vector<int>& nums3, vector<int>& nums4) {
        int count=0;
        unordered_map<int,int>mp;
        for(int i=0;i<nums1.size();i++){
            for(int j=0;j<nums2.size();j++){
                mp[nums1[i]+nums2[j]]++;
            }
        }
      
        for(int k=0;k<nums3.size();k++){
            for(int l=0;l<nums4.size();l++){
                count+=mp[-(nums3[k]+nums4[l])];
            }
        }
      
        return count;
    }
};