feat: add weekly contest 408

solution/3200-3299/3232.Find if Digit Game Can Be Won/README.md

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+---
+comments: true
+difficulty: 简单
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3200-3299/3232.Find%20if%20Digit%20Game%20Can%20Be%20Won/README.md
+---
+
+<!-- problem:start -->
+
+# [3232. 判断是否可以赢得数字游戏](https://leetcode.cn/problems/find-if-digit-game-can-be-won)
+
+[English Version](/solution/3200-3299/3232.Find%20if%20Digit%20Game%20Can%20Be%20Won/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给你一个 <strong>正整数 </strong>数组 <code>nums</code>。</p>
+
+<p>小红和小明正在玩游戏。在游戏中，小红可以从 <code>nums</code> 中选择所有个位数 <strong>或</strong> 所有两位数，剩余的数字归小明所有。如果小红所选数字之和 <strong>严格大于 </strong>小明的数字之和，则小红获胜。</p>
+
+<p>如果小红能赢得这场游戏，返回 <code>true</code>；否则，返回 <code>false</code>。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong><span class="example-io">nums = [1,2,3,4,10]</span></p>
+
+<p><strong>输出：</strong><span class="example-io">false</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>小红不管选个位数还是两位数都无法赢得比赛。</p>
+</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong><span class="example-io">nums = [1,2,3,4,5,14]</span></p>
+
+<p><strong>输出：</strong><span class="example-io">true</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>小红选择个位数可以赢得比赛，所选数字之和为 15。</p>
+</div>
+
+<p><strong class="example">示例 3：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong><span class="example-io">nums = [5,5,5,25]</span></p>
+
+<p><strong>输出：</strong><span class="example-io">true</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>小红选择两位数可以赢得比赛，所选数字之和为 25。</p>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 100</code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 99</code></li>
+</ul>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一：求和
+
+根据题目描述，只要个位数之和不等于两位数之和，那么小红一定可以选择一个较大的和来获胜。
+
+时间复杂度 $O(n)$，其中 $n$ 是数组 $\textit{nums}$ 的长度。空间复杂度 $O(1)$。
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def canAliceWin(self, nums: List[int]) -> bool:
+        a = sum(x for x in nums if x < 10)
+        b = sum(x for x in nums if x > 9)
+        return a != b
+```
+
+#### Java
+
+```java
+class Solution {
+    public boolean canAliceWin(int[] nums) {
+        int a = 0, b = 0;
+        for (int x : nums) {
+            if (x < 10) {
+                a += x;
+            } else {
+                b += x;
+            }
+        }
+        return a != b;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    bool canAliceWin(vector<int>& nums) {
+        int a = 0, b = 0;
+        for (int x : nums) {
+            if (x < 10) {
+                a += x;
+            } else {
+                b += x;
+            }
+        }
+        return a != b;
+    }
+};
+```
+
+#### Go
+
+```go
+func canAliceWin(nums []int) bool {
+	a, b := 0, 0
+	for _, x := range nums {
+		if x < 10 {
+			a += x
+		} else {
+			b += x
+		}
+	}
+	return a != b
+}
+```
+
+#### TypeScript
+
+```ts
+function canAliceWin(nums: number[]): boolean {
+    let [a, b] = [0, 0];
+    for (const x of nums) {
+        if (x < 10) {
+            a += x;
+        } else {
+            b += x;
+        }
+    }
+    return a !== b;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->

solution/3200-3299/3232.Find if Digit Game Can Be Won/README_EN.md

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+---
+comments: true
+difficulty: Easy
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3200-3299/3232.Find%20if%20Digit%20Game%20Can%20Be%20Won/README_EN.md
+---
+
+<!-- problem:start -->
+
+# [3232. Find if Digit Game Can Be Won](https://leetcode.com/problems/find-if-digit-game-can-be-won)
+
+[中文文档](/solution/3200-3299/3232.Find%20if%20Digit%20Game%20Can%20Be%20Won/README.md)
+
+## Description
+
+<!-- description:start -->
+
+<p>You are given an array of <strong>positive</strong> integers <code>nums</code>.</p>
+
+<p>Alice and Bob are playing a game. In the game, Alice can choose <strong>either</strong> all single-digit numbers or all double-digit numbers from <code>nums</code>, and the rest of the numbers are given to Bob. Alice wins if the sum of her numbers is <strong>strictly greater</strong> than the sum of Bob&#39;s numbers.</p>
+
+<p>Return <code>true</code> if Alice can win this game, otherwise, return <code>false</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [1,2,3,4,10]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">false</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>Alice cannot win by choosing either single-digit or double-digit numbers.</p>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [1,2,3,4,5,14]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">true</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>Alice can win by choosing single-digit numbers which have a sum equal to 15.</p>
+</div>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [5,5,5,25]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">true</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>Alice can win by choosing double-digit numbers which have a sum equal to 25.</p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 100</code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 99</code></li>
+</ul>
+
+<!-- description:end -->
+
+## Solutions
+
+<!-- solution:start -->
+
+### Solution 1: Summation
+
+According to the problem description, as long as the sum of the units digits is not equal to the sum of the tens digits, Xiaohong can always choose a larger sum to win.
+
+The time complexity is $O(n)$, where $n$ is the length of the array $\textit{nums}$. The space complexity is $O(1)$.
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def canAliceWin(self, nums: List[int]) -> bool:
+        a = sum(x for x in nums if x < 10)
+        b = sum(x for x in nums if x > 9)
+        return a != b
+```
+
+#### Java
+
+```java
+class Solution {
+    public boolean canAliceWin(int[] nums) {
+        int a = 0, b = 0;
+        for (int x : nums) {
+            if (x < 10) {
+                a += x;
+            } else {
+                b += x;
+            }
+        }
+        return a != b;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    bool canAliceWin(vector<int>& nums) {
+        int a = 0, b = 0;
+        for (int x : nums) {
+            if (x < 10) {
+                a += x;
+            } else {
+                b += x;
+            }
+        }
+        return a != b;
+    }
+};
+```
+
+#### Go
+
+```go
+func canAliceWin(nums []int) bool {
+	a, b := 0, 0
+	for _, x := range nums {
+		if x < 10 {
+			a += x
+		} else {
+			b += x
+		}
+	}
+	return a != b
+}
+```
+
+#### TypeScript
+
+```ts
+function canAliceWin(nums: number[]): boolean {
+    let [a, b] = [0, 0];
+    for (const x of nums) {
+        if (x < 10) {
+            a += x;
+        } else {
+            b += x;
+        }
+    }
+    return a !== b;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->

solution/3200-3299/3232.Find if Digit Game Can Be Won/Solution.cpp

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+class Solution {
+public:
+    bool canAliceWin(vector<int>& nums) {
+        int a = 0, b = 0;
+        for (int x : nums) {
+            if (x < 10) {
+                a += x;
+            } else {
+                b += x;
+            }
+        }
+        return a != b;
+    }
+};

solution/3200-3299/3232.Find if Digit Game Can Be Won/Solution.go

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+func canAliceWin(nums []int) bool {
+	a, b := 0, 0
+	for _, x := range nums {
+		if x < 10 {
+			a += x
+		} else {
+			b += x
+		}
+	}
+	return a != b
+}

solution/3200-3299/3232.Find if Digit Game Can Be Won/Solution.java

@@ -0,0 +1,13 @@
+class Solution {
+    public boolean canAliceWin(int[] nums) {
+        int a = 0, b = 0;
+        for (int x : nums) {
+            if (x < 10) {
+                a += x;
+            } else {
+                b += x;
+            }
+        }
+        return a != b;
+    }
+}

solution/3200-3299/3232.Find if Digit Game Can Be Won/Solution.py

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+class Solution:
+    def canAliceWin(self, nums: List[int]) -> bool:
+        a = sum(x for x in nums if x < 10)
+        b = sum(x for x in nums if x > 9)
+        return a != b

solution/3200-3299/3232.Find if Digit Game Can Be Won/Solution.ts

@@ -0,0 +1,11 @@
+function canAliceWin(nums: number[]): boolean {
+    let [a, b] = [0, 0];
+    for (const x of nums) {
+        if (x < 10) {
+            a += x;
+        } else {
+            b += x;
+        }
+    }
+    return a !== b;
+}