26_Maximum_Area_of_Longest_Diagonal_Rectangle.cpp

// 3000. Maximum Area of Longest Diagonal Rectangle

// You are given a 2D 0-indexed integer array dimensions.

// For all indices i, 0 <= i < dimensions.length, dimensions[i][0] represents the length and dimensions[i][1] represents the width of the rectangle i.

// Return the area of the rectangle having the longest diagonal. If there are multiple rectangles with the longest diagonal, return the area of the rectangle having the maximum area.

 

// Example 1:

// Input: dimensions = [[9,3],[8,6]]
// Output: 48
// Explanation: 
// For index = 0, length = 9 and width = 3. Diagonal length = sqrt(9 * 9 + 3 * 3) = sqrt(90) ≈ 9.487.
// For index = 1, length = 8 and width = 6. Diagonal length = sqrt(8 * 8 + 6 * 6) = sqrt(100) = 10.
// So, the rectangle at index 1 has a greater diagonal length therefore we return area = 8 * 6 = 48.
// Example 2:

// Input: dimensions = [[3,4],[4,3]]
// Output: 12
// Explanation: Length of diagonal is the same for both which is 5, so maximum area = 12.
 

// Constraints:

// 1 <= dimensions.length <= 100
// dimensions[i].length == 2
// 1 <= dimensions[i][0], dimensions[i][1] <= 100

class Solution
{
public:
    int areaOfMaxDiagonal(vector<vector<int>> &dimensions)
    {
        int maxArea = 0, Mdia2 = 0;
        for (auto &r : dimensions)
        {
            int w = r[0], h = r[1], dia2 = w * w + h * h;
            if (Mdia2 < dia2)
            {
                Mdia2 = dia2;
                maxArea = 0;
            }
            if (dia2 == Mdia2)
                maxArea = max(maxArea, w * h);
        }
        return maxArea;
    }
};