Easy C++ beats 98%
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\nEach comparison of rectangles defines a new rectangle in the overlap. We 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nEach comparison of rectangles defines a new rectangle in the overlap. We store the coordinates of the bottom left corner $(l_0,l_1)$ and the top right $(t_0,t_1)$. The largest square that can fit in this rectangle will have a side length that is the minimum of the vertical and horizontal distances. Calculate this in ```curr``` and update `ret` if we have found a new max side length. Return the square of `ret`.\n\n# Complexity\n- Time complexity: $$O(n^2)$$\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        long long ret = 0; \n        long long curr = 0;\n        long long l0, l1, t0, t1;\n\n        for (int i = 0; i < bottomLeft.size()-1; i++){\n            for (int j = i+1; j < topRight.size(); j++) {\n                l0 = max(bottomLeft[i][0], bottomLeft[j][0]);\n                l1 = max(bottomLeft[i][1], bottomLeft[j][1]);\n                t0 = min(topRight[i][0], topRight[j][0]);\n                t1 = min(topRight[i][1], topRight[j][1]);\n                curr = min(t0 -l0, t1-l1);\n                ret = curr > ret ? curr : ret;\n            }\n        }\n\n    return pow(ret, 2); \n    }\n};\n```
['C++']