index.ts

/**
 * # 3000. Maximum Area of Longest Diagonal Rectangle
 * @link [LeetCode](https://leetcode.com/problems/maximum-area-of-longest-diagonal-rectangle/description/)
 *
 * @description
 * 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.
 *
 * Constraints:
 * - 1 <= dimensions.length <= 100
 * - dimensions[i].length == 2
 * - 1 <= dimensions[i][0], dimensions[i][1] <= 100
 *
 * @example areaOfMaxDiagonal([[9,3],[8,6]]): 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 areaOfMaxDiagonal([[3,4],[4,3]]): 12
 * Explanation:
 * - Length of diagonal is the same for both which is 5, so maximum area = 12.
 *
 *
 * @param {number[][]} dimensions
 * @returns {number}
 */
function areaOfMaxDiagonal(dimensions: number[][]): number {
  let maxDiameter = 0;
  let maxArea = 0;
  for (const [length, width] of dimensions) {
    const diameter = Math.sqrt(length * length + width * width);
    if (maxDiameter < diameter) {
      maxDiameter = diameter;
      maxArea = length * width;
    } else if (maxDiameter === diameter) {
      maxArea = Math.max(maxArea, length * width);
    }
  }
  return maxArea;
}

export { areaOfMaxDiagonal };