BEATS 100% || EASY || DETAILED.
Intuition\n Describe your first thoughts on how to solve this problem. \n1. areaOfMaxDiagonal(int[][] dimensions)\n2. Takes a 2D array of dimensions as input.\n
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n1. areaOfMaxDiagonal(int[][] dimensions)\n2. Takes a 2D array of dimensions as input.\n3. Returns an integer representing the area of the desired rectangle.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n\n1. maxDiagonal stores the squared length of the longest diagonal found so far.\nmaxArea stores the maximum area found so far.\n2. **Iterating through rectangles**:\n\nThe code loops through each rectangle in the dimensions array:\nCalculates the squared length of the current rectangle\'s diagonal using the Pythagorean theorem (currentDiagonal).\n3. **Compares currentDiagonal with maxDiagonal**:\nIf currentDiagonal is greater, update maxDiagonal and maxArea with the current rectangle\'s values.\nIf currentDiagonal is equal to maxDiagonal, check if the current rectangle\'s area is greater than maxArea. If so, update maxArea.\n4. **Returning the result**:\n\nAfter the loop, maxArea holds the area of the desired rectangle, which is returned.\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\no(N)\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n  public int areaOfMaxDiagonal(int[][] dimensions) {\n    int maxDiagonalSQ = 0;\n    int maxArea = 0;\n    for (int i = 0; i < dimensions.length; i++) {\n      int currentDiagonal =(dimensions[i][0] * dimensions[i][0]) + (dimensions[i][1] * dimensions[i][1]);\n\n      if (currentDiagonal > maxDiagonalSQ || (currentDiagonal == maxDiagonalSQ && maxArea<dimensions[i][1] * dimensions[i][0])) {\n        maxDiagonalSQ = currentDiagonal;\n        maxArea = dimensions[i][1] * dimensions[i][0];\n      }\n    }\n\n    return maxArea;\n  }\n}\n\n```
['Java']