feat: add biweekly contest 173 (#4949)

Author: yanglbme

solution/1300-1399/1390.Four Divisors/README.md

@@ -198,11 +198,38 @@ function sumFourDivisors(nums: number[]): number {
         }
         return cnt === 4 ? s : 0;
     };
-    let ans = 0;
-    for (const x of nums) {
-        ans += f(x);
+    return nums.reduce((acc, x) => acc + f(x), 0);
+}
+```
+
+#### Rust
+
+```rust
+impl Solution {
+    pub fn sum_four_divisors(nums: Vec<i32>) -> i32 {
+        let f = |x: i32| -> i32 {
+            let mut cnt = 2;
+            let mut s = x + 1;
+            let mut i = 2;
+            while i <= x / i {
+                if x % i == 0 {
+                    cnt += 1;
+                    s += i;
+                    if i * i != x {
+                        cnt += 1;
+                        s += x / i;
+                    }
+                }
+                i += 1;
+            }
+            if cnt == 4 { s } else { 0 }
+        };
+        let mut ans = 0;
+        for x in nums {
+            ans += f(x);
+        }
+        ans
     }
-    return ans;
 }
 ```
 

solution/1300-1399/1390.Four Divisors/README_EN.md

@@ -27,7 +27,7 @@ tags:
 <pre>
 <strong>Input:</strong> nums = [21,4,7]
 <strong>Output:</strong> 32
-<strong>Explanation:</strong> 
+<strong>Explanation:</strong>
 21 has 4 divisors: 1, 3, 7, 21
 4 has 3 divisors: 1, 2, 4
 7 has 2 divisors: 1, 7
@@ -197,11 +197,38 @@ function sumFourDivisors(nums: number[]): number {
         }
         return cnt === 4 ? s : 0;
     };
-    let ans = 0;
-    for (const x of nums) {
-        ans += f(x);
+    return nums.reduce((acc, x) => acc + f(x), 0);
+}
+```
+
+#### Rust
+
+```rust
+impl Solution {
+    pub fn sum_four_divisors(nums: Vec<i32>) -> i32 {
+        let f = |x: i32| -> i32 {
+            let mut cnt = 2;
+            let mut s = x + 1;
+            let mut i = 2;
+            while i <= x / i {
+                if x % i == 0 {
+                    cnt += 1;
+                    s += i;
+                    if i * i != x {
+                        cnt += 1;
+                        s += x / i;
+                    }
+                }
+                i += 1;
+            }
+            if cnt == 4 { s } else { 0 }
+        };
+        let mut ans = 0;
+        for x in nums {
+            ans += f(x);
+        }
+        ans
     }
-    return ans;
 }
 ```
 

solution/1300-1399/1390.Four Divisors/Solution.rs

@@ -0,0 +1,26 @@
+impl Solution {
+    pub fn sum_four_divisors(nums: Vec<i32>) -> i32 {
+        let f = |x: i32| -> i32 {
+            let mut cnt = 2;
+            let mut s = x + 1;
+            let mut i = 2;
+            while i <= x / i {
+                if x % i == 0 {
+                    cnt += 1;
+                    s += i;
+                    if i * i != x {
+                        cnt += 1;
+                        s += x / i;
+                    }
+                }
+                i += 1;
+            }
+            if cnt == 4 { s } else { 0 }
+        };
+        let mut ans = 0;
+        for x in nums {
+            ans += f(x);
+        }
+        ans
+    }
+}

solution/1300-1399/1390.Four Divisors/Solution.ts

@@ -14,9 +14,5 @@ function sumFourDivisors(nums: number[]): number {
         }
         return cnt === 4 ? s : 0;
     };
-    let ans = 0;
-    for (const x of nums) {
-        ans += f(x);
-    }
-    return ans;
+    return nums.reduce((acc, x) => acc + f(x), 0);
 }

solution/3700-3799/3794.Reverse String Prefix/README.md

@@ -0,0 +1,137 @@
+---
+comments: true
+difficulty: 简单
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3794.Reverse%20String%20Prefix/README.md
+---
+
+<!-- problem:start -->
+
+# [3794. 反转字符串前缀](https://leetcode.cn/problems/reverse-string-prefix)
+
+[English Version](/solution/3700-3799/3794.Reverse%20String%20Prefix/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给你一个字符串 <code>s</code> 和一个整数 <code>k</code>。</p>
+
+<p>反转 <code>s</code> 的前 <code>k</code> 个字符，并返回结果字符串。</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">s = "abcd", k = 2</span></p>
+
+<p><strong>输出:</strong> <span class="example-io">"bacd"</span></p>
+
+<p><strong>解释:</strong></p>
+
+<p>前 <code>k = 2</code> 个字符 <code>"ab"</code> 反转为 <code>"ba"</code>。最终得到的结果字符串为 <code>"bacd"</code>。</p>
+</div>
+
+<p><strong class="example">示例 2:</strong></p>
+
+<div class="example-block">
+<p><strong>输入:</strong> <span class="example-io">s = "xyz", k = 3</span></p>
+
+<p><strong>输出:</strong> <span class="example-io">"zyx"</span></p>
+
+<p><strong>解释:</strong></p>
+
+<p>前 <code>k = 3</code> 个字符 <code>"xyz"</code> 反转为 <code>"zyx"</code>。最终得到的结果字符串为 <code>"zyx"</code>。</p>
+</div>
+
+<p><strong class="example">示例 3:</strong></p>
+
+<div class="example-block">
+<p><strong>输入:</strong> <span class="example-io">s = "hey", k = 1</span></p>
+
+<p><strong>输出:</strong> <span class="example-io">"hey"</span></p>
+
+<p><strong>解释:</strong></p>
+
+<p>前 <code>k = 1</code> 个字符 <code>"h"</code> 在反转后保持不变。最终得到的结果字符串为 <code>"hey"</code>。</p>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 100</code></li>
+	<li><code>s</code> 仅由小写英文字母组成。</li>
+	<li><code>1 &lt;= k &lt;= s.length</code></li>
+</ul>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一：模拟
+
+我们按照题目描述，将字符串的前 $k$ 个字符反转，然后与剩余的字符串拼接即可。
+
+时间复杂度 $O(n)$，空间复杂度 $O(n)$，其中 $n$ 是字符串的长度。
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def reversePrefix(self, s: str, k: int) -> str:
+        return s[:k][::-1] + s[k:]
+```
+
+#### Java
+
+```java
+class Solution {
+    public String reversePrefix(String s, int k) {
+        StringBuilder sb = new StringBuilder(s.substring(0, k));
+        return sb.reverse().toString() + s.substring(k);
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    string reversePrefix(string s, int k) {
+        string t = s.substr(0, k);
+        reverse(t.begin(), t.end());
+        return t + s.substr(k);
+    }
+};
+```
+
+#### Go
+
+```go
+func reversePrefix(s string, k int) string {
+	t := []byte(s[:k])
+	slices.Reverse(t)
+	return string(t) + s[k:]
+}
+```
+
+#### TypeScript
+
+```ts
+function reversePrefix(s: string, k: number): string {
+    return s.slice(0, k).split('').reverse().join('') + s.slice(k);
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->