Python one-liner O(n) time O(1) space
Intuition\nThe rectangle with the longest diagonal will have the longest w^2+l^2=diagonal^2.\nNot taking the square root on every rectangle will save some time 
# Intuition\nThe rectangle with the longest diagonal will have the longest w^2+l^2=diagonal^2.\nNot taking the square root on every rectangle will save some time (constant factor).\nThe rectangle with the maximum tuple (diagonal^2, area) can be found with one pass.\n\n# Approach\nUse built-in max, find the rectangle with the maximum diagonal/area.\n\n# Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(1)\n\n# Code\n```python3 []\nclass Solution:\n    def areaOfMaxDiagonal(self, dimensions: List[List[int]]) -> int:\n        return max(( (l**2+w**2,l*w) for l,w in dimensions))[1]\n\n```
['Python3']