c++||easy solution||
Complexity\n- Time complexity:\nO(N * N)\n\n# Code\n\nclass Solution {\npublic:\n    long long largestSquareArea(vector<vector<int>>& bottomLeft, vector<vector<
# Complexity\n- Time complexity:\nO(N * N)\n\n# Code\n```\nclass Solution {\npublic:\n    long long largestSquareArea(vector<vector<int>>& bottomLeft, vector<vector<int>>& topRight) {\n    long long maxarea = 0;\n    for(int i = 0; i < bottomLeft.size(); i++){\n        for(int j = i + 1; j < bottomLeft.size(); j++){\n            long long x1 = max(bottomLeft[i][0],bottomLeft[j][0]);\n            long long y1 = max(bottomLeft[i][1],bottomLeft[j][1]);\n            long long x2 = min(topRight[i][0],topRight[j][0]);\n            long long y2 = min(topRight[i][1],topRight[j][1]);\n            if(x1 < x2 && y1 < y2){\n                long long sq = min(x2 - x1,y2 - y1);\n                maxarea = max(maxarea,sq * sq);\n            }\n        }\n    }   \n    return maxarea;  \n    }\n};\n```
['Math', 'C++']