feat: add biweekly contest 135 and weekly contest 407

solution/1100-1199/1186.Maximum Subarray Sum with One Deletion/README.md

@@ -67,11 +67,11 @@ tags:
 
 ### 方法一：预处理 + 枚举
 
-我们可以先预处理出数组 $arr$ 以每个元素结尾和开头的最大子数组和，分别存入数组 $left$ 和 $right$ 中。
+我们可以先预处理出数组 $\textit{arr}$ 以每个元素结尾和开头的最大子数组和，分别存入数组 $\textit{left}$ 和 $\textit{right}$ 中。
 
-如果我们不删除任何元素，那么最大子数组和就是 $left[i]$ 或 $right[i]$ 中的最大值；如果我们删除一个元素，我们可以枚举 $[1..n-2]$ 中的每个位置 $i$，计算 $left[i-1] + right[i+1]$ 的值，取最大值即可。
+如果我们不删除任何元素，那么最大子数组和就是 $\textit{left}[i]$ 或 $\textit{right}[i]$ 中的最大值；如果我们删除一个元素，我们可以枚举 $[1..n-2]$ 中的每个位置 $i$，计算 $\textit{left}[i-1] + \textit{right}[i+1]$ 的值，取最大值即可。
 
-时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $arr$ 的长度。
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $\textit{arr}$ 的长度。
 
 <!-- tabs:start -->
 
@@ -195,6 +195,33 @@ function maximumSum(arr: number[]): number {
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn maximum_sum(arr: Vec<i32>) -> i32 {
+        let n = arr.len();
+        let mut left = vec![0; n];
+        let mut right = vec![0; n];
+        let mut s = 0;
+        for i in 0..n {
+            s = (s.max(0)) + arr[i];
+            left[i] = s;
+        }
+        s = 0;
+        for i in (0..n).rev() {
+            s = (s.max(0)) + arr[i];
+            right[i] = s;
+        }
+        let mut ans = *left.iter().max().unwrap();
+        for i in 1..n - 1 {
+            ans = ans.max(left[i - 1] + right[i + 1]);
+        }
+        ans
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/1100-1199/1186.Maximum Subarray Sum with One Deletion/README_EN.md

@@ -63,11 +63,11 @@ tags:
 
 ### Solution 1: Preprocessing + Enumeration
 
-We can first preprocess the array $arr$ to find the maximum subarray sum ending at and starting from each element, and store them in the arrays $left$ and $right$ respectively.
+We can preprocess the array $\textit{arr}$ to find the maximum subarray sum ending and starting with each element, storing them in arrays $\textit{left}$ and $\textit{right}$, respectively.
 
-If we do not delete any element, then the maximum subarray sum is the maximum value in $left[i]$ or $right[i]$. If we delete one element, we can enumerate each position $i$ in $[1..n-2]$, calculate the value of $left[i-1] + right[i+1]$, and take the maximum value.
+If we do not delete any element, then the maximum subarray sum is the maximum value in $\textit{left}[i]$ or $\textit{right}[i]$; if we delete an element, we can enumerate each position $i$ in $[1..n-2]$, calculate the value of $\textit{left}[i-1] + \textit{right}[i+1]$, and take the maximum value.
 
-The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $arr$.
+The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{arr}$.
 
 <!-- tabs:start -->
 
@@ -191,6 +191,33 @@ function maximumSum(arr: number[]): number {
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn maximum_sum(arr: Vec<i32>) -> i32 {
+        let n = arr.len();
+        let mut left = vec![0; n];
+        let mut right = vec![0; n];
+        let mut s = 0;
+        for i in 0..n {
+            s = (s.max(0)) + arr[i];
+            left[i] = s;
+        }
+        s = 0;
+        for i in (0..n).rev() {
+            s = (s.max(0)) + arr[i];
+            right[i] = s;
+        }
+        let mut ans = *left.iter().max().unwrap();
+        for i in 1..n - 1 {
+            ans = ans.max(left[i - 1] + right[i + 1]);
+        }
+        ans
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/1100-1199/1186.Maximum Subarray Sum with One Deletion/Solution.rs

@@ -0,0 +1,22 @@
+impl Solution {
+    pub fn maximum_sum(arr: Vec<i32>) -> i32 {
+        let n = arr.len();
+        let mut left = vec![0; n];
+        let mut right = vec![0; n];
+        let mut s = 0;
+        for i in 0..n {
+            s = (s.max(0)) + arr[i];
+            left[i] = s;
+        }
+        s = 0;
+        for i in (0..n).rev() {
+            s = (s.max(0)) + arr[i];
+            right[i] = s;
+        }
+        let mut ans = *left.iter().max().unwrap();
+        for i in 1..n - 1 {
+            ans = ans.max(left[i - 1] + right[i + 1]);
+        }
+        ans
+    }
+}

solution/3200-3299/3222.Find the Winning Player in Coin Game/README.md

@@ -0,0 +1,149 @@
+---
+comments: true
+difficulty: 简单
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3200-3299/3222.Find%20the%20Winning%20Player%20in%20Coin%20Game/README.md
+---
+
+<!-- problem:start -->
+
+# [3222. 求出硬币游戏的赢家](https://leetcode.cn/problems/find-the-winning-player-in-coin-game)
+
+[English Version](/solution/3200-3299/3222.Find%20the%20Winning%20Player%20in%20Coin%20Game/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给你两个 <strong>正</strong>&nbsp;整数&nbsp;<code>x</code>&nbsp;和&nbsp;<code>y</code>&nbsp;，分别表示价值为 75 和 10 的硬币的数目。</p>
+
+<p>Alice 和 Bob 正在玩一个游戏。每一轮中，Alice&nbsp;先进行操作，Bob 后操作。每次操作中，玩家需要拿出价值 <b>总和</b>&nbsp;为 115 的硬币。如果一名玩家无法执行此操作，那么这名玩家 <strong>输掉</strong>&nbsp;游戏。</p>
+
+<p>两名玩家都采取 <strong>最优</strong>&nbsp;策略，请你返回游戏的赢家。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b>x = 2, y = 7</span></p>
+
+<p><span class="example-io"><b>输出：</b>"Alice"</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>游戏一次操作后结束：</p>
+
+<ul>
+	<li>Alice 拿走 1 枚价值为 75 的硬币和 4 枚价值为 10 的硬币。</li>
+</ul>
+</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b>x = 4, y = 11</span></p>
+
+<p><span class="example-io"><b>输出：</b>"Bob"</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>游戏 2 次操作后结束：</p>
+
+<ul>
+	<li>Alice 拿走&nbsp;1 枚价值为 75 的硬币和 4 枚价值为 10 的硬币。</li>
+	<li>Bob 拿走&nbsp;1 枚价值为 75 的硬币和 4 枚价值为 10 的硬币。</li>
+</ul>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= x, y &lt;= 100</code></li>
+</ul>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一：数学
+
+由于每一轮的操作，会消耗 $2$ 枚价值为 $75$ 的硬币和 $8$ 枚价值为 $10$ 的硬币，因此，我们可以计算得到操作的轮数 $k = \min(x / 2, y / 8)$，然后更新 $x$ 和 $y$ 的值，此时 $x$ 和 $y$ 就是经过 $k$ 轮操作后剩余的硬币数目。
+
+如果 $x > 0$ 且 $y \geq 4$，那么 Alice 还可以继续操作，此时 Bob 就输了，返回 "Alice"；否则，返回 "Bob"。
+
+时间复杂度 $O(1)$，空间复杂度 $O(1)$。
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def losingPlayer(self, x: int, y: int) -> str:
+        k = min(x // 2, y // 8)
+        x -= k * 2
+        y -= k * 8
+        return "Alice" if x and y >= 4 else "Bob"
+```
+
+#### Java
+
+```java
+class Solution {
+    public String losingPlayer(int x, int y) {
+        int k = Math.min(x / 2, y / 8);
+        x -= k * 2;
+        y -= k * 8;
+        return x > 0 && y >= 4 ? "Alice" : "Bob";
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    string losingPlayer(int x, int y) {
+        int k = min(x / 2, y / 8);
+        x -= k * 2;
+        y -= k * 8;
+        return x && y >= 4 ? "Alice" : "Bob";
+    }
+};
+```
+
+#### Go
+
+```go
+func losingPlayer(x int, y int) string {
+	k := min(x/2, y/8)
+	x -= 2 * k
+	y -= 8 * k
+	if x > 0 && y >= 4 {
+		return "Alice"
+	}
+	return "Bob"
+}
+```
+
+#### TypeScript
+
+```ts
+function losingPlayer(x: number, y: number): string {
+    const k = Math.min((x / 2) | 0, (y / 8) | 0);
+    x -= k * 2;
+    y -= k * 8;
+    return x && y >= 4 ? 'Alice' : 'Bob';
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->