Easy intuitive C++ Solution
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```\n#define ll long long\nclass Solution {\npublic:\n    long long largestSquareArea(vector<vector<int>>& bottomLeft, vector<vector<int>>& topRight) {\n        int n = bottomLeft.size();\n        ll ans = 0;\n\n        for(int i = 0; i < n; i++){\n            for(int j = i + 1; j < n; j++){\n                // first rectangle\n                int ax1 = bottomLeft[i][0];\n                int ay1 = bottomLeft[i][1];\n                int ax2 = topRight[i][0];\n                int ay2 = topRight[i][1];\n\n                // second rectangle\n                int bx1 = bottomLeft[j][0];\n                int by1 = bottomLeft[j][1];\n                int bx2 = topRight[j][0];\n                int by2 = topRight[j][1];\n\n                int left = max(ax1, bx1);\n                int right = min(ax2, bx2);\n                int top = min(ay2, by2);\n                int down = max(ay1, by1);\n\n                // overlapping\n                if(right > left && top > down){\n                    int width = right - left;\n                    int height = top - down;\n\n                    int mini = min(width, height);\n                    ans = max(ans, (ll)((ll)mini * (ll)mini));\n                }\n            }\n        }\n\n        return ans;\n    }\n};\n```
['C++']