Simple solution using unordered map | c++
Take an unordered map and simply store the values with their number of occurences in the array. Now check if the second value (ie number of occurences) stored i
Take an unordered map and simply store the values with their number of occurences in the array. Now check if the second value (ie number of occurences) stored in the map is equal to the size of vector (ie nums) then it implies that the value corresponding to this is present in every array. So just push this value in the vector \'v\'. The finally made vector v will be the answer.\n\n```\n vector<int> intersection(vector<vector<int>>& nums) {\n        \n        int n=nums.size();\n        unordered_map<int, int> mp;\n        vector<int> v;\n        \n        for(int i=0; i<n; i++){\n            for(int j=0; j<nums[i].size(); j++){\n                mp[nums[i][j]]++;\n            }\n        }\n        \n        for(auto x: mp){\n            if(x.second == n){\n                v.push_back(x.first);\n            }\n        }\n        \n        sort(v.begin(), v.end());\n        \n        return v;\n    }\n\t\n\t```
['C']