Hashmap Solution ||  Easy to understand C++
The idea is to traverse through complete 2d array and count the frequency of every element and at last if the size of outer array is equal to the frequency we r
The idea is to traverse through complete 2d array and count the frequency of every element and at last if the size of outer array is equal to the frequency we return the value.\n\n\uD83D\uDE42\n\n```\nclass Solution {\npublic:\n    vector<int> intersection(vector<vector<int>>& nums) {\n        map<int,int> hm;\n        int n = 0;\n        for(auto it: nums){\n            for(auto x: it){\n                hm[x]++;\n            }\n            n++;\n        }\n        vector<int> ans;\n        for(auto it: hm){\n            if(it.second == n){\n                ans.push_back(it.first);\n            }\n        }\n        return ans;\n    }\n};\n```
['C']