c++, O(n^2)
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```cpp []\nclass Solution {\npublic:\n    long long largestSquareArea(vector<vector<int>>& bottomLeft, vector<vector<int>>& topRight) {\n        ///min(x2i,x2j) - max (x1i,x1j) | \n        int n = bottomLeft.size();\n        long long ans = 0;\n        for(int i = 0;i<n-1;i++){\n            for(int j = i+1;j<n;j++){\n                int x1i = bottomLeft[i][0],y1i = bottomLeft[i][1],x2i = topRight[i][0],y2i = topRight[i][1];\n                int x1j = bottomLeft[j][0],y1j = bottomLeft[j][1],x2j = topRight[j][0],y2j = topRight[j][1];\n                if(x1i>x1j){\n                    swap(x1i,x1j); swap(y1i,y1j); swap(x2i,x2j); swap(y2i,y2j);\n                }\n                int minX = max(x1i,x1j),maxX = min(x2i,x2j);\n                if(minX >= maxX) continue;\n                int minY,maxY;\n                if(y1j <= y1i){\n                    minY = min(y2j,y1i);\n                    maxY = min(y2i,y2j);\n                }\n                else{\n                    minY = min(y1j,y2i);\n                    maxY = min(y2j,y2i);\n                }\n                if(minY >= maxY) continue;\n                int len = min((maxY-minY),(maxX-minX));\n              //  cout<<i<<" "<<j<<" "<<len<<" "<<minX<<" "<<maxX<<" "<<minY<<" "<<maxY<<endl;\n                ans = max(ans,1LL*len*len);\n            }\n        }\n        return ans;\n    }\n};\n```
['C++']