Package g2901_3000.s3000_maximum_area_of_longest_diagonal_rectangle

Class Solution

  • All Implemented Interfaces:
    
public final class Solution

    3000 - Maximum Area of Longest Diagonal Rectangle.

    Easy

    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

    • Nested Class Summary

      Nested Classes 
      Modifier and Type Class Description

    • Field Summary

      Fields 
      Modifier and Type Field Description

    • Constructor Summary

      Constructors 
      Constructor Description
      Solution()

    • Enum Constant Summary

      Enum Constants 
      Enum Constant Description

    • Method Summary

      Modifier and Type Method Description
      final Integer areaOfMaxDiagonal(Array<IntArray> dimensions)

      • Methods inherited from class java.lang.Object

        clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait