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Intuition\nTo solve the problem of finding the area of the rectangle with the longest diagonal, first, consider that the diagonal length of a rectangle is calcu
# Intuition\nTo solve the problem of finding the area of the rectangle with the longest diagonal, first, consider that the diagonal length of a rectangle is calculated using the Pythagorean theorem. Given a list of rectangles with their dimensions, our goal is to compute the diagonal length for each rectangle and determine which rectangle has the longest diagonal. If multiple rectangles have the same diagonal length, we need to select the one with the largest area.\n\n# Approach\n1. **Initialize Variables**: Use variables to keep track of the maximum diagonal length (`diag`) and the area corresponding to this diagonal length (`area`).\n\n2. **Iterate Through Rectangles**: For each rectangle, compute the diagonal length using the formula:\n   \\[\n   \\text{Diagonal} = \\sqrt{\\text{length}^2 + \\text{width}^2}\n   \\]\n\n3. **Update Maximum Diagonal and Area**:\n   - If the current diagonal is greater than the previously recorded maximum diagonal (`diag`), update both `diag` and `area` with the current rectangle\'s diagonal and area.\n   - If the diagonal is equal to the maximum diagonal, compare the current rectangle\'s area with the previously recorded maximum area and update if necessary.\n\n4. **Return Result**: After processing all rectangles, return the area of the rectangle with the longest diagonal. If there are multiple rectangles with the same longest diagonal, return the one with the maximum area.\n\n# Complexity\n- Time complexity: \\(O(n)\\) \n  - Each rectangle is processed once to compute the diagonal and potentially update the maximum diagonal and area. Therefore, the time complexity is linear with respect to the number of rectangles.\n\n- Space complexity: \\(O(1)\\)\n  - The algorithm uses a constant amount of extra space for variables (diagonal length, area) regardless of the number of rectangles. Hence, the space complexity is constant.\n\n# Code\n```\nclass Solution:\n    def areaOfMaxDiagonal(self, dimensions: List[List[int]]) -> int:\n        diag = 0\n        area = 0\n\n        for length,width in dimensions:\n            diagonal = sqrt(length ** 2 + width ** 2)\n            if diag <= diagonal:\n                if diag == diagonal:\n                    area = max(area,length * width)\n                else:\n                    area = length * width\n                    diag = diagonal\n\n                \n        return area\n        \n```
['Array', 'Python3']