Simple python3 solution
Complexity\n- Time complexity: O(n^2)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(n)\n Add your space complexity here, e.g. O(n) \n\n# 
# Complexity\n- Time complexity: $$O(n^2)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n``` python3 []\nclass Solution:\n    def largestSquareArea(self, bottomLeft: List[List[int]], topRight: List[List[int]]) -> int:\n        data = sorted(\n            (x, y, u, v)\n            for (x, y), (u, v) in zip(bottomLeft, topRight)\n        )\n\n        result = 0\n        for i, (x, y, u, v) in enumerate(data):\n            for j in range(i + 1, len(data)):\n                X, Y, U, V = data[j]\n                if U < X:\n                    break\n                side = min(min(u, U) - max(x, X), min(v, V) - max(y, Y))\n                result = max(result, side)\n        return result ** 2\n\n```
['Geometry', 'Sorting', 'Python3']