Java : Beats 66%
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 {\n    public long largestSquareArea(int[][] bottomLeft, int[][] topRight) {\n        int n = bottomLeft.length;\n        long maxArea = 0;\n        for(int i=0;i<n;i++){\n            int ax = bottomLeft[i][0];\n            int ay = bottomLeft[i][1];\n\n            int bx = topRight[i][0];\n            int by = topRight[i][1];\n\n            for(int j=i+1;j<n;j++){\n                int cx = bottomLeft[j][0];\n                int cy = bottomLeft[j][1];\n\n                int dx = topRight[j][0];\n                int dy = topRight[j][1];\n\n                int x1 = Math.max(ax,cx);\n                int y1 = Math.max(ay,cy);\n                int x2 = Math.min(bx,dx);\n                int y2 = Math.min(by,dy);\n\n                int minSide = Math.min(x2 - x1,y2 - y1);\n                long area = (long) Math.pow(Math.max(minSide,0),2);\n                maxArea = Math.max(maxArea,area);\n            }\n        }\n        return maxArea;\n    }\n}\n```
['Java']