Easy Solution | C++
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: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    long long largestSquareArea(vector<vector<int>>& bottomLeft, vector<vector<int>>& topRight) {\n        int n=bottomLeft.size();\n        long long int ans=0,ma=0;\n        for(int i=0;i<n;i++){\n            for(int j=i+1;j<n;j++){\n    long long int x1=max(bottomLeft[j][0],bottomLeft[i][0]);\n     long long int y1=max(bottomLeft[j][1],bottomLeft[i][1]);\n                \n        long long int x2=min(topRight[j][0],topRight[i][0]);\n     long long int y2=min(topRight[j][1],topRight[i][1]);\n                \n        //\n         if(x2-x1>0 && y2-y1>0)ans=min(x2-x1,y2-y1);\n                else ans=0;\n    \n     ma=max(ma,ans);\n\n     \n \n            }\n            \n        }\n        long long int z=ma*ma;\n        return z;\n        \n    }\n};\n```
['C++']