3251 Maximum Area Of Longest Diagonal Rectangle

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

use std::cmp::max;

impl Solution {
    pub fn area_of_max_diagonal(dimensions: Vec<Vec<i32>>) -> i32 {
        let mut area = 0;
        let mut diag = 0;
        let mut prev = 0;

        for k in dimensions.clone() {
            diag = max(k[0].pow(2) + k[1].pow(2), diag);
        }

        for k in dimensions {
            if diag == k[0].pow(2) + k[1].pow(2) {
                area = max(area, k[0] * k[1]);
            }
        }

        return area;
    }
}