Today I have done leetcode's August LeetCoding Challenge with cpp.

August LeetCoding Challenge 26

Description

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

Solution

class Solution {
public:
  int areaOfMaxDiagonal(vector<vector<int>>& dimensions) {
    int longestDiagonal = 0;
    int answer = 0;
    for(const auto &d : dimensions) {
      int dia = d[0] * d[0] + d[1] * d[1];
      if(dia > longestDiagonal) {
        answer = d[0] * d[1];
        longestDiagonal = dia;
      } else if(dia == longestDiagonal) {
        answer = max(answer, d[0] * d[1]);
      }
    }
    return answer;
  }
};

// Accepted
// 816/816 cases passed (0 ms)
// Your runtime beats 100 % of cpp submissions
// Your memory usage beats 77.05 % of cpp submissions (29.2 MB)