README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/3000-3099/3000.Maximum%20Area%20of%20Longest%20Diagonal%20Rectangle/README_EN.md	1249	Weekly Contest 379 Q1	Array

3000. Maximum Area of Longest Diagonal Rectangle

中文文档

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

Python3

class Solution:
    def areaOfMaxDiagonal(self, dimensions: List[List[int]]) -> int:
        ans = mx = 0
        for l, w in dimensions:
            t = l**2 + w**2
            if mx < t:
                mx = t
                ans = l * w
            elif mx == t:
                ans = max(ans, l * w)
        return ans

Java

class Solution {
    public int areaOfMaxDiagonal(int[][] dimensions) {
        int ans = 0, mx = 0;
        for (var d : dimensions) {
            int l = d[0], w = d[1];
            int t = l * l + w * w;
            if (mx < t) {
                mx = t;
                ans = l * w;
            } else if (mx == t) {
                ans = Math.max(ans, l * w);
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int areaOfMaxDiagonal(vector<vector<int>>& dimensions) {
        int ans = 0, mx = 0;
        for (auto& d : dimensions) {
            int l = d[0], w = d[1];
            int t = l * l + w * w;
            if (mx < t) {
                mx = t;
                ans = l * w;
            } else if (mx == t) {
                ans = max(ans, l * w);
            }
        }
        return ans;
    }
};

Go

func areaOfMaxDiagonal(dimensions [][]int) (ans int) {
	mx := 0
	for _, d := range dimensions {
		l, w := d[0], d[1]
		t := l*l + w*w
		if mx < t {
			mx = t
			ans = l * w
		} else if mx == t {
			ans = max(ans, l*w)
		}
	}
	return
}

TypeScript

function areaOfMaxDiagonal(dimensions: number[][]): number {
    let [ans, mx] = [0, 0];
    for (const [l, w] of dimensions) {
        const t = l * l + w * w;
        if (mx < t) {
            mx = t;
            ans = l * w;
        } else if (mx === t) {
            ans = Math.max(ans, l * w);
        }
    }
    return ans;
}