simple | easy
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    O(n^2)\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        long long maxi = 0;\n        bool inter = false;\n        int n = bottomLeft.size();\n        int m = topRight.size();\n        for(int i=0; i<n; i++){\n            for(int j=i+1; j<n; j++){\n                int x1= max(bottomLeft[i][0], bottomLeft[j][0]);\n                int y1= max(bottomLeft[i][1], bottomLeft[j][1]);\n                int x2= min(topRight[i][0],   topRight[j][0]);\n                int y2= min(topRight[i][1],   topRight[j][1]);\n                \n                if(x1<x2 && y1<y2){\n                    inter = true;\n                    long long side = min(x2-x1, y2-y1);\n                    maxi = max(maxi, side*side);\n                }\n            }\n        }\n        if(!inter)\n            return 0;\n        return maxi;\n    }\n};\n```
['C++']