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Commit e13c2c2

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feat: add biweekly contest 135 and weekly contest 407 (doocs#3297)
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‎solution/1100-1199/1186.Maximum Subarray Sum with One Deletion/README.md‎

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### 方法一:预处理 + 枚举
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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。
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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。
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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,ドル计算 $left[i-1] + right[i+1]$ 的值,取最大值即可。
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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,ドル计算 $\textit{left}[i-1] + \textit{right}[i+1]$ 的值,取最大值即可。
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时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 为数组 $arr$ 的长度。
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时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 为数组 $\textit{arr}$ 的长度。
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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 maximum_sum(arr: Vec<i32>) -> i32 {
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let n = arr.len();
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let mut left = vec![0; n];
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let mut right = vec![0; n];
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let mut s = 0;
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for i in 0..n {
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s = (s.max(0)) + arr[i];
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left[i] = s;
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}
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s = 0;
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for i in (0..n).rev() {
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s = (s.max(0)) + arr[i];
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right[i] = s;
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}
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let mut ans = *left.iter().max().unwrap();
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for i in 1..n - 1 {
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ans = ans.max(left[i - 1] + right[i + 1]);
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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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<!-- tabs:end -->
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<!-- solution:end -->

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

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### Solution 1: Preprocessing + Enumeration
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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.
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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},ドル respectively.
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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],ドル calculate the value of $left[i-1] + right[i+1],ドル and take the maximum value.
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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],ドル calculate the value of $\textit{left}[i-1] + \textit{right}[i+1],ドル and take the maximum value.
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The time complexity is $O(n),ドル and the space complexity is $O(n)$. Here, $n$ is the length of the array $arr$.
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The time complexity is $O(n),ドル and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{arr}$.
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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 maximum_sum(arr: Vec<i32>) -> i32 {
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let n = arr.len();
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let mut left = vec![0; n];
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let mut right = vec![0; n];
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let mut s = 0;
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for i in 0..n {
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s = (s.max(0)) + arr[i];
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left[i] = s;
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}
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s = 0;
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for i in (0..n).rev() {
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s = (s.max(0)) + arr[i];
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right[i] = s;
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}
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let mut ans = *left.iter().max().unwrap();
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for i in 1..n - 1 {
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ans = ans.max(left[i - 1] + right[i + 1]);
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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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<!-- tabs:end -->
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<!-- solution:end -->
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impl Solution {
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pub fn maximum_sum(arr: Vec<i32>) -> i32 {
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let n = arr.len();
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let mut left = vec![0; n];
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let mut right = vec![0; n];
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let mut s = 0;
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for i in 0..n {
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s = (s.max(0)) + arr[i];
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left[i] = s;
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}
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s = 0;
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for i in (0..n).rev() {
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s = (s.max(0)) + arr[i];
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right[i] = s;
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}
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let mut ans = *left.iter().max().unwrap();
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for i in 1..n - 1 {
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ans = ans.max(left[i - 1] + right[i + 1]);
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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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comments: true
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difficulty: 简单
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3200-3299/3222.Find%20the%20Winning%20Player%20in%20Coin%20Game/README.md
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---
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<!-- problem:start -->
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# [3222. 求出硬币游戏的赢家](https://leetcode.cn/problems/find-the-winning-player-in-coin-game)
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[English Version](/solution/3200-3299/3222.Find%20the%20Winning%20Player%20in%20Coin%20Game/README_EN.md)
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## 题目描述
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<!-- description:start -->
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<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>
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<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>
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<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>
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<p>&nbsp;</p>
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<p><strong class="example">示例 1:</strong></p>
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<div class="example-block">
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<p><span class="example-io"><b>输入:</b>x = 2, y = 7</span></p>
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<p><span class="example-io"><b>输出:</b>"Alice"</span></p>
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<p><strong>解释:</strong></p>
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<p>游戏一次操作后结束:</p>
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<ul>
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<li>Alice 拿走 1 枚价值为 75 的硬币和 4 枚价值为 10 的硬币。</li>
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</ul>
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</div>
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<p><strong class="example">示例 2:</strong></p>
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<div class="example-block">
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<p><span class="example-io"><b>输入:</b>x = 4, y = 11</span></p>
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<p><span class="example-io"><b>输出:</b>"Bob"</span></p>
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<p><strong>解释:</strong></p>
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<p>游戏 2 次操作后结束:</p>
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<ul>
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<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>
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<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>
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</ul>
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</div>
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<p>&nbsp;</p>
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<p><strong>提示:</strong></p>
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<ul>
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<li><code>1 &lt;= x, y &lt;= 100</code></li>
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</ul>
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<!-- description:end -->
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## 解法
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<!-- solution:start -->
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### 方法一:数学
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由于每一轮的操作,会消耗 2ドル$ 枚价值为 75ドル$ 的硬币和 8ドル$ 枚价值为 10ドル$ 的硬币,因此,我们可以计算得到操作的轮数 $k = \min(x / 2, y / 8),ドル然后更新 $x$ 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。
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如果 $x > 0$ 且 $y \geq 4,ドル那么 Alice 还可以继续操作,此时 Bob 就输了,返回 "Alice";否则,返回 "Bob"。
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时间复杂度 $O(1),ドル空间复杂度 $O(1)$。
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<!-- tabs:start -->
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#### Python3
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```python
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class Solution:
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def losingPlayer(self, x: int, y: int) -> str:
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k = min(x // 2, y // 8)
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x -= k * 2
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y -= k * 8
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return "Alice" if x and y >= 4 else "Bob"
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```
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#### Java
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```java
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class Solution {
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public String losingPlayer(int x, int y) {
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int k = Math.min(x / 2, y / 8);
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x -= k * 2;
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y -= k * 8;
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return x > 0 && y >= 4 ? "Alice" : "Bob";
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}
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}
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```
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#### C++
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```cpp
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class Solution {
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public:
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string losingPlayer(int x, int y) {
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int k = min(x / 2, y / 8);
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x -= k * 2;
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y -= k * 8;
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return x && y >= 4 ? "Alice" : "Bob";
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}
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};
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```
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#### Go
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```go
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func losingPlayer(x int, y int) string {
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k := min(x/2, y/8)
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x -= 2 * k
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y -= 8 * k
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if x > 0 && y >= 4 {
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return "Alice"
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}
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return "Bob"
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}
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```
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#### TypeScript
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```ts
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function losingPlayer(x: number, y: number): string {
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const k = Math.min((x / 2) | 0, (y / 8) | 0);
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x -= k * 2;
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y -= k * 8;
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return x && y >= 4 ? 'Alice' : 'Bob';
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

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