Simple c++ n^2 solution
Complexity\n- Time complexity: O(n.n)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(1)\n Add your space complexity here, e.g. O(n) \n\n# 
# Complexity\n- Time complexity: O(n.n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n    \n    long long solveImpl(\n        int bottomX, int bottomY, int topX, int topY, \n        int bottomX2, int bottomY2, int topX2, int topY2\n    ){\n        long long mini = min(\n            (min(topX, topX2) - max(bottomX, bottomX2)), \n            (min(topY, topY2) - max(bottomY, bottomY2))\n        );\n        mini = max(0ll, mini);\n        return mini * mini;\n    }\n    \n    long long solve(vector<vector<int>>& bottomLeft, vector<vector<int>>& topRight, int i, int j){\n        return solveImpl(\n            bottomLeft[i][0], bottomLeft[i][1], topRight[i][0], topRight[i][1], \n            bottomLeft[j][0], bottomLeft[j][1], topRight[j][0], topRight[j][1]\n        );\n    }\n    \n    long long largestSquareArea(vector<vector<int>>& bottomLeft, vector<vector<int>>& topRight) {\n        int n = bottomLeft.size();\n        long long ans = 0;\n        for(int i = 0; i < n; i++){\n            for(int j = i + 1; j < n; j++){\n                ans = max(ans, solve(bottomLeft, topRight, i, j));\n            }\n        }\n        return ans;\n    }\n};\n```
['C++']