Math + logic + O(n^2) complexity
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        long long int side = 0;\n        long long areaans=0;\n        long long area = 0;\n        for(int i=0;i<bottomLeft.size();i++){\n            for(int j=i+1;j<bottomLeft.size();j++){\n                if(bottomLeft[j][1]< topRight[i][1] && bottomLeft[j][0]< topRight[i][0]\n                   &&bottomLeft[i][1]<topRight[j][1] &&bottomLeft[i][0]<topRight[j][0] )\n                {\n                   long long int bottomlx,bottomly,toprx,topry;\n                    bottomlx = max(bottomLeft[i][0],bottomLeft[j][0]);\n                    bottomly = max(bottomLeft[i][1],bottomLeft[j][1]);\n\n                    toprx = min(topRight[i][0],topRight[j][0]);\n                    topry = min(topRight[i][1],topRight[j][1]);\n\n                    side = min(abs(toprx-bottomlx), abs(bottomly-topry));\n\n                    area = side*side;\n                }\n                areaans = max(areaans,area);\n            }\n        }\n        return areaans;\n    }\n};\n```
['C++']