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Commit 1eb9f7f

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feat: add solutions to lc problem: No.3000 (#4669)
1 parent ee147f7 commit 1eb9f7f

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4 files changed

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-3
lines changed

4 files changed

+141
-3
lines changed

‎solution/3000-3099/3000.Maximum Area of Longest Diagonal Rectangle/README.md‎

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<!-- solution:start -->
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### 方法一
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### 方法一:数学
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根据勾股定理,矩形的对角线的平方为 $l^2 + w^2,ドル其中 $l$ 和 $w$ 分别是矩形的长度和宽度。
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我们可以遍历所有矩形,计算它们的对角线长度的平方,并记录下最大的对角线长度和对应的面积。
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遍历结束后,我们返回记录的最大面积。
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时间复杂度 $O(n),ドル其中 $n$ 是矩形的数量。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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@@ -161,6 +169,50 @@ function areaOfMaxDiagonal(dimensions: number[][]): number {
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}
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```
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#### Rust
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```rust
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impl Solution {
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pub fn area_of_max_diagonal(dimensions: Vec<Vec<i32>>) -> i32 {
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let mut ans = 0;
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let mut mx = 0;
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for d in dimensions {
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let l = d[0];
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let w = d[1];
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let t = l * l + w * w;
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if mx < t {
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mx = t;
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ans = l * w;
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} else if mx == t {
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ans = ans.max(l * w);
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}
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}
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ans
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}
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}
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```
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#### C#
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```cs
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public class Solution {
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public int AreaOfMaxDiagonal(int[][] dimensions) {
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int ans = 0, mx = 0;
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foreach (var d in dimensions) {
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int l = d[0], w = d[1];
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int t = l * l + w * w;
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if (mx < t) {
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mx = t;
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ans = l * w;
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} else if (mx == t) {
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ans = Math.Max(ans, l * w);
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}
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}
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return ans;
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->

‎solution/3000-3099/3000.Maximum Area of Longest Diagonal Rectangle/README_EN.md‎

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<pre>
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<strong>Input:</strong> dimensions = [[9,3],[8,6]]
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<strong>Output:</strong> 48
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<strong>Explanation:</strong>
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<strong>Explanation:</strong>
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For index = 0, length = 9 and width = 3. Diagonal length = sqrt(9 * 9 + 3 * 3) = sqrt(90) &asymp;<!-- notionvc: 882cf44c-3b17-428e-9c65-9940810216f1 --> 9.487.
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For index = 1, length = 8 and width = 6. Diagonal length = sqrt(8 * 8 + 6 * 6) = sqrt(100) = 10.
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So, the rectangle at index 1 has a greater diagonal length therefore we return area = 8 * 6 = 48.
@@ -59,7 +59,15 @@ So, the rectangle at index 1 has a greater diagonal length therefore we return a
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<!-- solution:start -->
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### Solution 1
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### Solution 1: Mathematics
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According to the Pythagorean theorem, the square of the diagonal of a rectangle is $l^2 + w^2,ドル where $l$ and $w$ are the length and width of the rectangle, respectively.
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We can iterate through all the rectangles, calculate the square of their diagonal lengths, and keep track of the maximum diagonal length and the corresponding area.
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After the iteration, we return the recorded maximum area.
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The time complexity is $O(n),ドル where $n$ is the number of rectangles. The space complexity is $O(1)$.
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<!-- tabs:start -->
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}
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```
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#### Rust
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```rust
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impl Solution {
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pub fn area_of_max_diagonal(dimensions: Vec<Vec<i32>>) -> i32 {
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let mut ans = 0;
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let mut mx = 0;
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for d in dimensions {
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let l = d[0];
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let w = d[1];
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let t = l * l + w * w;
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if mx < t {
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mx = t;
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ans = l * w;
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} else if mx == t {
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ans = ans.max(l * w);
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}
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}
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ans
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}
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}
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```
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#### C#
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```cs
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public class Solution {
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public int AreaOfMaxDiagonal(int[][] dimensions) {
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int ans = 0, mx = 0;
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foreach (var d in dimensions) {
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int l = d[0], w = d[1];
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int t = l * l + w * w;
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if (mx < t) {
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mx = t;
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ans = l * w;
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} else if (mx == t) {
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ans = Math.Max(ans, l * w);
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}
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}
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return ans;
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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public class Solution {
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public int AreaOfMaxDiagonal(int[][] dimensions) {
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int ans = 0, mx = 0;
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foreach (var d in dimensions) {
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int l = d[0], w = d[1];
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int t = l * l + w * w;
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if (mx < t) {
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mx = t;
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ans = l * w;
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} else if (mx == t) {
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ans = Math.Max(ans, l * w);
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}
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}
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return ans;
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}
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}
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impl Solution {
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pub fn area_of_max_diagonal(dimensions: Vec<Vec<i32>>) -> i32 {
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let mut ans = 0;
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let mut mx = 0;
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for d in dimensions {
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let l = d[0];
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let w = d[1];
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let t = l * l + w * w;
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if mx < t {
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mx = t;
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ans = l * w;
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} else if mx == t {
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ans = ans.max(l * w);
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}
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}
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ans
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}
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}

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