Simple math 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```\nclass Solution {\npublic:\n    long long largestSquareArea(vector<vector<int>>& bottomLeft, vector<vector<int>>& topRight) {\n        long long maxarea = 0;\n        int n = bottomLeft.size();\n        for (int i = 0; i < n; i++) {\n            int x1 = bottomLeft[i][0];\n            int x2 = topRight[i][0];\n            int y1 = bottomLeft[i][1];\n            int y2 = topRight[i][1];\n            for (int j = i+1; j < n; j++) {\n                int x11 = bottomLeft[j][0];\n                int x22 = topRight[j][0];\n                int y11 = bottomLeft[j][1];\n                int y22 = topRight[j][1];\n                \n                // Check if there is an overlap\n                if (x1 > x22 || y1 > y22 || x2 < x11 || y2 < y11) continue;\n\n                // Calculate the sides of the overlapping rectangle\n                int l = max(x1, x11) - min(x2, x22);\n                int b = max(y1, y11) - min(y2, y22);\n\n                // Ensure l and b are non-negative\n                l = abs(l);\n                b = abs(b);\n\n                // Calculate the largest possible square area within the overlap\n                long long int area = min(l, b) * (long long)min(l, b);\n                maxarea = max(maxarea, area);\n            }\n        }\n        return maxarea;\n    }\n};\n\n```
['Math', 'C++']