Merge branch 'doocs:main' into main

Author: taoyq1988

solution/2300-2399/2364.Count Number of Bad Pairs/README.md

@@ -63,15 +63,15 @@ tags:
 
 ### 方法一：式子转换 + 哈希表
 
-根据题目描述，我们可以得知，对于任意的 $i \lt j$，如果 $j - i \neq nums[j] - nums[i]$，则 $(i, j)$ 是一个坏数对。
+根据题目描述，我们可以得知，对于任意的 $i \lt j$，如果 $j - i \neq \textit{nums}[j] - \textit{nums}[i]$，则 $(i, j)$ 是一个坏数对。
 
-我们可以将式子转换为 $i - nums[i] \neq j - nums[j]$。这启发我们用哈希表 $cnt$ 来统计 $i - nums[i]$ 的出现次数。
+我们可以将式子转换为 $i - \textit{nums}[i] \neq j - \textit{nums}[j]$。这启发我们用哈希表 $cnt$ 来统计 $i - \textit{nums}[i]$ 的出现次数。
 
-我们遍历数组，对于当前元素 $nums[i]$，我们将 $i - cnt[i - nums[i]]$ 加到答案中，然后将 $i - nums[i]$ 的出现次数加 $1$。
+遍历数组，对于当前元素 $\textit{nums}[i]$，我们将 $i - cnt[i - \textit{nums}[i]]$ 加到答案中，然后将 $i - \textit{nums}[i]$ 的出现次数加 $1$。
 
-最终，我们返回答案即可。
+最后，我们返回答案即可。
 
-时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组的长度。
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $\textit{nums}$ 的长度。
 
 <!-- tabs:start -->
 
@@ -97,8 +97,7 @@ class Solution {
         long ans = 0;
         for (int i = 0; i < nums.length; ++i) {
             int x = i - nums[i];
-            ans += i - cnt.getOrDefault(x, 0);
-            cnt.merge(x, 1, Integer::sum);
+            ans += i - cnt.merge(x, 1, Integer::sum) + 1;
         }
         return ans;
     }
@@ -115,8 +114,7 @@ public:
         long long ans = 0;
         for (int i = 0; i < nums.size(); ++i) {
             int x = i - nums[i];
-            ans += i - cnt[x];
-            ++cnt[x];
+            ans += i - cnt[x]++;
         }
         return ans;
     }
@@ -152,6 +150,26 @@ function countBadPairs(nums: number[]): number {
 }
 ```
 
+#### Rust
+
+```rust
+use std::collections::HashMap;
+
+impl Solution {
+    pub fn count_bad_pairs(nums: Vec<i32>) -> i64 {
+        let mut cnt: HashMap<i32, i64> = HashMap::new();
+        let mut ans: i64 = 0;
+        for (i, &num) in nums.iter().enumerate() {
+            let x = i as i32 - num;
+            let count = *cnt.get(&x).unwrap_or(&0);
+            ans += i as i64 - count;
+            *cnt.entry(x).or_insert(0) += 1;
+        }
+        ans
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/2300-2399/2364.Count Number of Bad Pairs/README_EN.md

@@ -63,15 +63,15 @@ There are a total of 5 bad pairs, so we return 5.
 
 ### Solution 1: Equation Transformation + Hash Table
 
-From the problem description, we know that for any $i < j$, if $j - i \neq nums[j] - nums[i]$, then $(i, j)$ is a bad pair.
+According to the problem description, for any $i \lt j$, if $j - i \neq \textit{nums}[j] - \textit{nums}[i]$, then $(i, j)$ is a bad pair.
 
-We can transform the equation to $i - nums[i] \neq j - nums[j]$. This inspires us to use a hash table $cnt$ to count the occurrences of $i - nums[i]$.
+We can transform the equation into $i - \textit{nums}[i] \neq j - \textit{nums}[j]$. This suggests using a hash table $cnt$ to count the occurrences of $i - \textit{nums}[i]$.
 
-We iterate through the array. For the current element $nums[i]$, we add $i - cnt[i - nums[i]]$ to the answer, then increment the count of $i - nums[i]$ by $1$.
+While iterating through the array, for the current element $\textit{nums}[i]$, we add $i - cnt[i - \textit{nums}[i]]$ to the answer, and then increment the count of $i - \textit{nums}[i]$ by $1$.
 
 Finally, we return the answer.
 
-The time complexity is $O(n)$ and the space complexity is $O(n)$, where $n$ is the length of the array.
+The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the array $\textit{nums}$.
 
 <!-- tabs:start -->
 
@@ -97,8 +97,7 @@ class Solution {
         long ans = 0;
         for (int i = 0; i < nums.length; ++i) {
             int x = i - nums[i];
-            ans += i - cnt.getOrDefault(x, 0);
-            cnt.merge(x, 1, Integer::sum);
+            ans += i - cnt.merge(x, 1, Integer::sum) + 1;
         }
         return ans;
     }
@@ -115,8 +114,7 @@ public:
         long long ans = 0;
         for (int i = 0; i < nums.size(); ++i) {
             int x = i - nums[i];
-            ans += i - cnt[x];
-            ++cnt[x];
+            ans += i - cnt[x]++;
         }
         return ans;
     }
@@ -152,6 +150,26 @@ function countBadPairs(nums: number[]): number {
 }
 ```
 
+#### Rust
+
+```rust
+use std::collections::HashMap;
+
+impl Solution {
+    pub fn count_bad_pairs(nums: Vec<i32>) -> i64 {
+        let mut cnt: HashMap<i32, i64> = HashMap::new();
+        let mut ans: i64 = 0;
+        for (i, &num) in nums.iter().enumerate() {
+            let x = i as i32 - num;
+            let count = *cnt.get(&x).unwrap_or(&0);
+            ans += i as i64 - count;
+            *cnt.entry(x).or_insert(0) += 1;
+        }
+        ans
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/2300-2399/2364.Count Number of Bad Pairs/Solution.cpp

@@ -5,8 +5,7 @@ class Solution {
         long long ans = 0;
         for (int i = 0; i < nums.size(); ++i) {
             int x = i - nums[i];
-            ans += i - cnt[x];
-            ++cnt[x];
+            ans += i - cnt[x]++;
         }
         return ans;
     }

solution/2300-2399/2364.Count Number of Bad Pairs/Solution.java

@@ -4,8 +4,7 @@ public long countBadPairs(int[] nums) {
         long ans = 0;
         for (int i = 0; i < nums.length; ++i) {
             int x = i - nums[i];
-            ans += i - cnt.getOrDefault(x, 0);
-            cnt.merge(x, 1, Integer::sum);
+            ans += i - cnt.merge(x, 1, Integer::sum) + 1;
         }
         return ans;
     }

solution/2300-2399/2364.Count Number of Bad Pairs/Solution.rs

@@ -0,0 +1,15 @@
+use std::collections::HashMap;
+
+impl Solution {
+    pub fn count_bad_pairs(nums: Vec<i32>) -> i64 {
+        let mut cnt: HashMap<i32, i64> = HashMap::new();
+        let mut ans: i64 = 0;
+        for (i, &num) in nums.iter().enumerate() {
+            let x = i as i32 - num;
+            let count = *cnt.get(&x).unwrap_or(&0);
+            ans += i as i64 - count;
+            *cnt.entry(x).or_insert(0) += 1;
+        }
+        ans
+    }
+}

solution/3500-3599/3506.Find Time Required to Eliminate Bacterial Strains/README.md

@@ -2,6 +2,11 @@
 comments: true
 difficulty: 困难
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3506.Find%20Time%20Required%20to%20Eliminate%20Bacterial%20Strains/README.md
+tags:
+    - 贪心
+    - 数组
+    - 数学
+    - 堆（优先队列）
 ---
 
 <!-- problem:start -->
@@ -24,9 +29,9 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3506.Fi
 
 <ul>
 	<li>第 <code>i</code> 个细菌菌株需要 <code>timeReq[i]</code> 个时间单位来被消除。</li>
-	<li>单个白细胞只能消除 <strong>一种</strong> 细菌菌株。之后，白细胞耗尽，无法执行任何其他任务。</li>
+	<li>单个白细胞只能消除 <strong>一个</strong> 细菌菌株。之后，白细胞耗尽，无法执行任何其他任务。</li>
 	<li>一个白细胞可以将自身分裂为两个白细胞，但这需要&nbsp;<code>splitTime</code>&nbsp;单位时间。一旦分裂，两个白细胞就可以 <strong>并行</strong> 消灭细菌。</li>
-	<li>仅有一个白细胞可以针对一个单一细菌菌株工作。多个白细胞不能同时攻击一个菌株。</li>
+	<li>一个白细胞仅可以攻击一个细菌菌株。多个白细胞不能同时攻击一个菌株。</li>
 </ul>
 
 <p>您必须确定消除所有细菌菌株所需的 <strong>最短</strong> 时间。</p>
@@ -58,7 +63,7 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3506.Fi
 <div class="example-block">
 <p><span class="example-io"><b>输入：</b>timeReq = [10,4], splitTime = 5</span></p>
 
-<p><b>输出：</b>5</p>
+<p><b>输出：</b>15</p>
 
 <p><strong>解释：</strong></p>
 

solution/3500-3599/3506.Find Time Required to Eliminate Bacterial Strains/README_EN.md

@@ -2,6 +2,11 @@
 comments: true
 difficulty: Hard
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3506.Find%20Time%20Required%20to%20Eliminate%20Bacterial%20Strains/README_EN.md
+tags:
+    - Greedy
+    - Array
+    - Math
+    - Heap (Priority Queue)
 ---
 
 <!-- problem:start -->

solution/3500-3599/3507.Minimum Pair Removal to Sort Array I/README.md

@@ -2,6 +2,14 @@
 comments: true
 difficulty: 简单
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3507.Minimum%20Pair%20Removal%20to%20Sort%20Array%20I/README.md
+tags:
+    - 数组
+    - 哈希表
+    - 链表
+    - 双向链表
+    - 有序集合
+    - 模拟
+    - 堆（优先队列）
 ---
 
 <!-- problem:start -->

solution/3500-3599/3507.Minimum Pair Removal to Sort Array I/README_EN.md

@@ -2,6 +2,14 @@
 comments: true
 difficulty: Easy
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3507.Minimum%20Pair%20Removal%20to%20Sort%20Array%20I/README_EN.md
+tags:
+    - Array
+    - Hash Table
+    - Linked List
+    - Doubly-Linked List
+    - Ordered Set
+    - Simulation
+    - Heap (Priority Queue)
 ---
 
 <!-- problem:start -->

solution/3500-3599/3508.Implement Router/README.md

@@ -2,6 +2,13 @@
 comments: true
 difficulty: 中等
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3508.Implement%20Router/README.md
+tags:
+    - 设计
+    - 队列
+    - 数组
+    - 哈希表
+    - 二分查找
+    - 有序集合
 ---
 
 <!-- problem:start -->

solution/3500-3599/3508.Implement Router/README_EN.md

@@ -2,6 +2,13 @@
 comments: true
 difficulty: Medium
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3508.Implement%20Router/README_EN.md
+tags:
+    - Design
+    - Queue
+    - Array
+    - Hash Table
+    - Binary Search
+    - Ordered Set
 ---
 
 <!-- problem:start -->

solution/3500-3599/3509.Maximum Product of Subsequences With an Alternating Sum Equal to K/README.md

@@ -2,6 +2,10 @@
 comments: true
 difficulty: 困难
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3509.Maximum%20Product%20of%20Subsequences%20With%20an%20Alternating%20Sum%20Equal%20to%20K/README.md
+tags:
+    - 数组
+    - 哈希表
+    - 动态规划
 ---
 
 <!-- problem:start -->

solution/3500-3599/3509.Maximum Product of Subsequences With an Alternating Sum Equal to K/README_EN.md

@@ -2,6 +2,10 @@
 comments: true
 difficulty: Hard
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3509.Maximum%20Product%20of%20Subsequences%20With%20an%20Alternating%20Sum%20Equal%20to%20K/README_EN.md
+tags:
+    - Array
+    - Hash Table
+    - Dynamic Programming
 ---
 
 <!-- problem:start -->

solution/3500-3599/3512.Minimum Operations to Make Array Sum Divisible by K/README.md

@@ -2,6 +2,9 @@
 comments: true
 difficulty: 简单
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3512.Minimum%20Operations%20to%20Make%20Array%20Sum%20Divisible%20by%20K/README.md
+tags:
+    - 数组
+    - 数学
 ---
 
 <!-- problem:start -->
@@ -84,32 +87,60 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3512.Mi
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：求和取模
+
+题目实际上是求数组元素之和对 $k$ 取模的结果。因此，我们只需要遍历数组，计算出所有元素之和，然后对 $k$ 取模，最后返回这个结果即可。
+
+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 minOperations(self, nums: List[int], k: int) -> int:
+        return sum(nums) % k
 ```
 
 #### Java
 
 ```java
-
+class Solution {
+    public int minOperations(int[] nums, int k) {
+        return Arrays.stream(nums).sum() % k;
+    }
+}
 ```
 
 #### C++
 
 ```cpp
-
+class Solution {
+public:
+    int minOperations(vector<int>& nums, int k) {
+        return reduce(nums.begin(), nums.end(), 0) % k;
+    }
+};
 ```
 
 #### Go
 
 ```go
+func minOperations(nums []int, k int) (ans int) {
+	for _, x := range nums {
+		ans = (ans + x) % k
+	}
+	return
+}
+```
+
+#### TypeScript
 
+```ts
+function minOperations(nums: number[], k: number): number {
+    return nums.reduce((acc, x) => acc + x, 0) % k;
+}
 ```
 
 <!-- tabs:end -->