good beat %
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n    long long largestSquareArea(vector<vector<int>>& bottomLeft, vector<vector<int>>& topRight) {\n        int n = bottomLeft.size();\n        int m = topRight.size();\n        long long ans = 0;\n        for(int i = 0; i < n; i++) {\n            for(int j = i + 1; j < n; j++) {\n                int maxx = max(bottomLeft[i][0], bottomLeft[j][0]);\n                int maxy = max(bottomLeft[i][1], bottomLeft[j][1]);\n                int minx = min(topRight[i][0], topRight[j][0]);\n                int miny = min(topRight[i][1], topRight[j][1]);\n                if(maxx < minx && maxy < miny) {\n                    int x = min((minx - maxx), (miny - maxy));\n                    ans = max(ans, static_cast<long long>(x) * x);\n                }\n            }\n        }\n        return ans;\n    }\n};\n\n```
['Array', 'Math', 'Geometry', 'C++']