C++ Naive Implementation
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 maxArea {0};\n        for (size_t i = 0; i < bottomLeft.size(); ++i) {\n            for (size_t j = i + 1; j < bottomLeft.size(); ++j) {\n                if (bottomLeft[i][0] >= topRight[j][0] || topRight[i][1] <= bottomLeft[j][1] ||\n                    topRight[i][0] <= bottomLeft[j][0] || bottomLeft[i][1] >= topRight[j][1]) \n                    continue;\n                int right = std::min(topRight[i][0], topRight[j][0]);\n                int left = std::max(bottomLeft[i][0], bottomLeft[j][0]);\n                int top = std::min(topRight[i][1], topRight[j][1]);\n                int bottom = std::max(bottomLeft[i][1], bottomLeft[j][1]);\n                int w = right - left >= top - bottom ? top - bottom : right - left;\n                maxArea = std::max(maxArea, 1LL * w * w);\n            }\n        }\n        return maxArea;\n        \n    }\n};\n```
['C++']