3000. Maximum Area of Longest Diagonal Rectangle
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.
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
Solutions
Solution 1
1 2 3 4 5 6 7 8 9 10 11 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 |
|