feat: update lc problems

solution/2600-2699/2610.Convert an Array Into a 2D Array With Conditions/README.md

@@ -68,13 +68,13 @@ nums 中的所有元素都有用到，并且每一行都由不同的整数组成
 
 ### 方法一：数组或哈希表
 
-我们先用数组或哈希表 $cnt$ 统计数组 $nums$ 中每个元素出现的次数。
+我们先用一个数组或者哈希表 $\textit{cnt}$ 统计数组 $\textit{nums}$ 中每个元素出现的次数。
 
-然后遍历 $cnt$，对于每个元素 $x$，我们将其添加到答案列表中的第 $0$ 行，第 $1$ 行，第 $2$ 行，...，第 $cnt[x]-1$ 行。
+然后遍历 $\textit{cnt}$，对于每个元素 $x$，我们将其添加到答案列表中的第 $0$ 行，第 $1$ 行，第 $2$ 行，...，第 $\textit{cnt}[x]-1$ 行。
 
 最后返回答案列表即可。
 
-时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $nums$ 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 -->
 
@@ -171,7 +171,7 @@ func findMatrix(nums []int) (ans [][]int) {
 function findMatrix(nums: number[]): number[][] {
     const ans: number[][] = [];
     const n = nums.length;
-    const cnt: number[] = new Array(n + 1).fill(0);
+    const cnt: number[] = Array(n + 1).fill(0);
     for (const x of nums) {
         ++cnt[x];
     }
@@ -187,6 +187,33 @@ function findMatrix(nums: number[]): number[][] {
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn find_matrix(nums: Vec<i32>) -> Vec<Vec<i32>> {
+        let n = nums.len();
+        let mut cnt = vec![0; n + 1];
+        let mut ans: Vec<Vec<i32>> = Vec::new();
+
+        for &x in &nums {
+            cnt[x as usize] += 1;
+        }
+
+        for x in 1..=n as i32 {
+            for j in 0..cnt[x as usize] {
+                if ans.len() <= j {
+                    ans.push(Vec::new());
+                }
+                ans[j].push(x);
+            }
+        }
+
+        ans
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/2600-2699/2610.Convert an Array Into a 2D Array With Conditions/README_EN.md

@@ -68,13 +68,13 @@ It can be shown that we cannot have less than 3 rows in a valid array.</pre>
 
 ### Solution 1: Array or Hash Table
 
-We use an array or hash table $cnt$ to count the number of occurrences of each element in the array $nums$.
+We first use an array or hash table $\textit{cnt}$ to count the frequency of each element in the array $\textit{nums}$.
 
-Then we traverse the $cnt$ array, add $x$ to the $0$th row, the $1$st row, the $2$nd row, ..., the ($cnt[x]-1$)th row of the answer list.
+Then we iterate through $\textit{cnt}$. For each element $x$, we add it to the 0th row, 1st row, 2nd row, ..., and $(cnt[x]-1)$th row of the answer list.
 
-Finally, return the answer list.
+Finally, we return the answer list.
 
-The time complexity is $O(n)$ and the space complexity is $O(n)$, where $n$ is the length of the array $nums$.
+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 -->
 
@@ -171,7 +171,7 @@ func findMatrix(nums []int) (ans [][]int) {
 function findMatrix(nums: number[]): number[][] {
     const ans: number[][] = [];
     const n = nums.length;
-    const cnt: number[] = new Array(n + 1).fill(0);
+    const cnt: number[] = Array(n + 1).fill(0);
     for (const x of nums) {
         ++cnt[x];
     }
@@ -187,6 +187,33 @@ function findMatrix(nums: number[]): number[][] {
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn find_matrix(nums: Vec<i32>) -> Vec<Vec<i32>> {
+        let n = nums.len();
+        let mut cnt = vec![0; n + 1];
+        let mut ans: Vec<Vec<i32>> = Vec::new();
+
+        for &x in &nums {
+            cnt[x as usize] += 1;
+        }
+
+        for x in 1..=n as i32 {
+            for j in 0..cnt[x as usize] {
+                if ans.len() <= j {
+                    ans.push(Vec::new());
+                }
+                ans[j].push(x);
+            }
+        }
+
+        ans
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/2600-2699/2610.Convert an Array Into a 2D Array With Conditions/Solution.rs

@@ -0,0 +1,22 @@
+impl Solution {
+    pub fn find_matrix(nums: Vec<i32>) -> Vec<Vec<i32>> {
+        let n = nums.len();
+        let mut cnt = vec![0; n + 1];
+        let mut ans: Vec<Vec<i32>> = Vec::new();
+
+        for &x in &nums {
+            cnt[x as usize] += 1;
+        }
+
+        for x in 1..=n as i32 {
+            for j in 0..cnt[x as usize] {
+                if ans.len() <= j {
+                    ans.push(Vec::new());
+                }
+                ans[j].push(x);
+            }
+        }
+
+        ans
+    }
+}

solution/2600-2699/2610.Convert an Array Into a 2D Array With Conditions/Solution.ts

@@ -1,7 +1,7 @@
 function findMatrix(nums: number[]): number[][] {
     const ans: number[][] = [];
     const n = nums.length;
-    const cnt: number[] = new Array(n + 1).fill(0);
+    const cnt: number[] = Array(n + 1).fill(0);
     for (const x of nums) {
         ++cnt[x];
     }

solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/README.md

@@ -41,7 +41,7 @@ tags:
 <strong>输入：</strong>nums = [1,10,3,4,19]
 <strong>输出：</strong>133
 <strong>解释：</strong>下标三元组 (1, 2, 4) 的值是 (nums[1] - nums[2]) * nums[4] = 133 。
-可以证明不存在值大于 133 的有序下标三元组。
+可以证明不存在值大于 133 的有序下标三元组。 
 </pre>
 
 <p><strong class="example">示例 3：</strong></p>

solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/README_EN.md

@@ -31,7 +31,7 @@ tags:
 <strong>Input:</strong> nums = [12,6,1,2,7]
 <strong>Output:</strong> 77
 <strong>Explanation:</strong> The value of the triplet (0, 2, 4) is (nums[0] - nums[2]) * nums[4] = 77.
-It can be shown that there are no ordered triplets of indices with a value greater than 77.
+It can be shown that there are no ordered triplets of indices with a value greater than 77. 
 </pre>
 
 <p><strong class="example">Example 2:</strong></p>

solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/README.md

@@ -41,7 +41,7 @@ tags:
 <strong>输入：</strong>nums = [1,10,3,4,19]
 <strong>输出：</strong>133
 <strong>解释：</strong>下标三元组 (1, 2, 4) 的值是 (nums[1] - nums[2]) * nums[4] = 133 。
-可以证明不存在值大于 133 的有序下标三元组。
+可以证明不存在值大于 133 的有序下标三元组。 
 </pre>
 
 <p><strong class="example">示例 3：</strong></p>

solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/README_EN.md

@@ -31,7 +31,7 @@ tags:
 <strong>Input:</strong> nums = [12,6,1,2,7]
 <strong>Output:</strong> 77
 <strong>Explanation:</strong> The value of the triplet (0, 2, 4) is (nums[0] - nums[2]) * nums[4] = 77.
-It can be shown that there are no ordered triplets of indices with a value greater than 77.
+It can be shown that there are no ordered triplets of indices with a value greater than 77. 
 </pre>
 
 <p><strong class="example">Example 2:</strong></p>

solution/3400-3499/3481.Apply Substitutions/README.md

@@ -14,68 +14,70 @@ tags:
 
 <!-- problem:start -->
 
-# [3481. Apply Substitutions 🔒](https://leetcode.cn/problems/apply-substitutions)
+# [3481. 应用替换 🔒](https://leetcode.cn/problems/apply-substitutions)
 
 [English Version](/solution/3400-3499/3481.Apply%20Substitutions/README_EN.md)
 
 ## 题目描述
 
 <!-- description:start -->
 
-<p data-end="384" data-start="34">You are given a <code>replacements</code> mapping and a <code>text</code> string that may contain <strong>placeholders</strong> formatted as <code data-end="139" data-start="132">%var%</code>, where each <code>var</code> corresponds to a key in the <code>replacements</code> mapping. Each replacement value may itself contain <strong>one or more</strong> such <strong>placeholders</strong>. Each <strong>placeholder</strong> is replaced by the value associated with its corresponding replacement key.</p>
+<p data-end="384" data-start="34">给定一个&nbsp;<code>replacements</code>&nbsp;映射和一个可能包含格式为 <code>%var%</code> <strong>占位符&nbsp;</strong>的字符串&nbsp;<code>text</code>，其中每个&nbsp;<code>var</code>&nbsp;对应&nbsp;<code>replacements</code>&nbsp;中的一个键。每个替换值本身可能包含 <strong>一个或多个</strong> 此类<strong>占位符</strong>。每个 <strong>占位符</strong> 都被与其相应的替换键对应的值替换。</p>
 
-<p data-end="353" data-start="34">Return the fully substituted <code>text</code> string which <strong>does not</strong> contain any <strong>placeholders</strong>.</p>
+<p data-end="353" data-start="34">返回完全替换后 <strong>不</strong> 含任何 <strong>占位符</strong> 的&nbsp;<code>text</code>&nbsp;字符串。</p>
 
 <p>&nbsp;</p>
-<p><strong class="example">Example 1:</strong></p>
+
+<p><strong class="example">示例 1：</strong></p>
 
 <div class="example-block">
-<p><strong>Input:</strong> <span class="example-io">replacements = [[&quot;A&quot;,&quot;abc&quot;],[&quot;B&quot;,&quot;def&quot;]], text = &quot;%A%_%B%&quot;</span></p>
+<p><span class="example-io"><b>输入：</b>replacements = [["A","abc"],["B","def"]], text = "%A%_%B%"</span></p>
 
-<p><strong>Output:</strong> <span class="example-io">&quot;abc_def&quot;</span></p>
+<p><strong>输出：</strong><span class="example-io">"abc_def"</span></p>
 
-<p><strong>Explanation:</strong></p>
+<p><strong>解释：</strong></p>
 
 <ul data-end="238" data-start="71">
-	<li data-end="138" data-start="71">The mapping associates <code data-end="101" data-start="96">&quot;A&quot;</code> with <code data-end="114" data-start="107">&quot;abc&quot;</code> and <code data-end="124" data-start="119">&quot;B&quot;</code> with <code data-end="137" data-start="130">&quot;def&quot;</code>.</li>
-	<li data-end="203" data-start="139">Replace <code data-end="154" data-start="149">%A%</code> with <code data-end="167" data-start="160">&quot;abc&quot;</code> and <code data-end="177" data-start="172">%B%</code> with <code data-end="190" data-start="183">&quot;def&quot;</code> in the text.</li>
-	<li data-end="238" data-start="204">The final text becomes <code data-end="237" data-start="226">&quot;abc_def&quot;</code>.</li>
+	<li data-end="138" data-start="71">映射将&nbsp;<code data-end="101" data-start="96">"A"</code> 与&nbsp;<code data-end="114" data-start="107">"abc"</code>&nbsp;关联，并将&nbsp;<code data-end="124" data-start="119">"B"</code> 与&nbsp;<code data-end="137" data-start="130">"def"</code>&nbsp;关联。</li>
+	<li data-end="203" data-start="139">用&nbsp;<code data-end="167" data-start="160">"abc"</code>&nbsp;替换文本中的&nbsp;<code data-end="154" data-start="149">%A%</code>，并用&nbsp;<code data-end="190" data-start="183">"def"</code>&nbsp;替换文本中的&nbsp;<code data-end="177" data-start="172">%B%</code>。</li>
+	<li data-end="238" data-start="204">最终文本变为&nbsp;<code data-end="237" data-start="226">"abc_def"</code>。</li>
 </ul>
 </div>
 
-<p><strong class="example">Example 2:</strong></p>
+<p><strong class="example">示例 2：</strong></p>
 
 <div class="example-block">
-<p><strong>Input:</strong> <span class="example-io">replacements = [[&quot;A&quot;,&quot;bce&quot;],[&quot;B&quot;,&quot;ace&quot;],[&quot;C&quot;,&quot;abc%B%&quot;]], text = &quot;%A%_%B%_%C%&quot;</span></p>
+<p><span class="example-io"><b>输入：</b>replacements = [["A","bce"],["B","ace"],["C","abc%B%"]], text = "%A%_%B%_%C%"</span></p>
 
-<p><strong>Output:</strong> <span class="example-io">&quot;bce_ace_abcace&quot;</span></p>
+<p><span class="example-io"><b>输出：</b>"bce_ace_abcace"</span></p>
 
-<p><strong>Explanation:</strong></p>
+<p><strong>解释：</strong></p>
 
 <ul data-end="541" data-is-last-node="" data-is-only-node="" data-start="255">
-	<li data-end="346" data-start="255">The mapping associates <code data-end="285" data-start="280">&quot;A&quot;</code> with <code data-end="298" data-start="291">&quot;bce&quot;</code>, <code data-end="305" data-start="300">&quot;B&quot;</code> with <code data-end="318" data-start="311">&quot;ace&quot;</code>, and <code data-end="329" data-start="324">&quot;C&quot;</code> with <code data-end="345" data-start="335">&quot;abc%B%&quot;</code>.</li>
-	<li data-end="411" data-start="347">Replace <code data-end="362" data-start="357">%A%</code> with <code data-end="375" data-start="368">&quot;bce&quot;</code> and <code data-end="385" data-start="380">%B%</code> with <code data-end="398" data-start="391">&quot;ace&quot;</code> in the text.</li>
-	<li data-end="496" data-start="412">Then, for <code data-end="429" data-start="424">%C%</code>, substitute <code data-end="447" data-start="442">%B%</code> in <code data-end="461" data-start="451">&quot;abc%B%&quot;</code> with <code data-end="474" data-start="467">&quot;ace&quot;</code> to obtain <code data-end="495" data-start="485">&quot;abcace&quot;</code>.</li>
-	<li data-end="541" data-is-last-node="" data-start="497">The final text becomes <code data-end="540" data-start="522">&quot;bce_ace_abcace&quot;</code>.</li>
+	<li data-end="346" data-start="255">映射将&nbsp;<code data-end="285" data-start="280">"A"</code> 与&nbsp;<code data-end="298" data-start="291">"bce"</code>&nbsp;关联，<code data-end="305" data-start="300">"B"</code> 与&nbsp;<code data-end="318" data-start="311">"ace"</code>&nbsp;关联，以及&nbsp;<code data-end="329" data-start="324">"C"</code> 与&nbsp;<code data-end="345" data-start="335">"abc%B%"</code>&nbsp;关联。</li>
+	<li data-end="411" data-start="347">用&nbsp;<code data-end="375" data-start="368">"bce"</code>&nbsp;替换文本中的&nbsp;<code data-end="362" data-start="357">%A%</code>，并用&nbsp;<code data-end="398" data-start="391">"ace"</code>&nbsp;替换文本中的&nbsp;<code data-end="385" data-start="380">%B%</code>。</li>
+	<li data-end="496" data-start="412">接着，对于&nbsp;<code data-end="429" data-start="424">%C%</code>，用&nbsp;<code data-end="474" data-start="467">"ace"</code> 替换&nbsp;<code data-end="461" data-start="451">"abc%B%"</code>&nbsp;中的&nbsp;<code data-end="447" data-start="442">%B%</code>&nbsp;来得到&nbsp;<code data-end="495" data-start="485">"abcace"</code>。</li>
+	<li data-end="541" data-is-last-node="" data-start="497">最终文本变为&nbsp;<code data-end="540" data-start="522">"bce_ace_abcace"</code>。</li>
 </ul>
 </div>
 
 <p>&nbsp;</p>
-<p><strong>Constraints:</strong></p>
+
+<p><strong>提示：</strong></p>
 
 <ul>
 	<li data-end="1432" data-start="1398"><code>1 &lt;= replacements.length &lt;= 10</code></li>
-	<li data-end="1683" data-start="1433">Each element of <code data-end="1465" data-start="1451">replacements</code> is a two-element list <code data-end="1502" data-start="1488">[key, value]</code>, where:
+	<li data-end="1683" data-start="1433"><code data-end="1465" data-start="1451">replacements</code>&nbsp;中的每个元素是一个双值列表&nbsp;<code data-end="1502" data-start="1488">[key, value]</code>，其中：
 	<ul data-end="1683" data-start="1513">
-		<li data-end="1558" data-start="1513"><code data-end="1520" data-start="1515">key</code> is a single uppercase English letter.</li>
-		<li data-end="1683" data-start="1561"><code data-end="1570" data-start="1563">value</code> is a non-empty string of at most 8 characters that may contain zero or more placeholders formatted as <code data-end="1682" data-start="1673">%&lt;key&gt;%</code>.</li>
+		<li data-end="1558" data-start="1513"><code data-end="1520" data-start="1515">key</code>&nbsp;是一个大写英语字母。</li>
+		<li data-end="1683" data-start="1561"><code data-end="1570" data-start="1563">value</code>&nbsp;是一个最多有 8 个字符，可能包含 0 个或更多格式为&nbsp;<code data-end="1682" data-start="1673">%&lt;key&gt;%</code> 的占位符的非空字符串。</li>
 	</ul>
 	</li>
-	<li data-end="726" data-start="688">All replacement keys are unique.</li>
-	<li data-end="1875" data-start="1723">The <code>text</code> string is formed by concatenating all key placeholders (formatted as <code data-end="1808" data-start="1799">%&lt;key&gt;%</code>) randomly from the replacements mapping, separated by underscores.</li>
+	<li data-end="726" data-start="688">所有的替换键互不相同。</li>
+	<li data-end="1875" data-start="1723"><code>text</code>&nbsp;字符串是通过从替换映射中随机串联所有 key 占位符（格式为 <code>%&lt;key&gt;%</code>）而形成的，以虚线分隔。</li>
 	<li data-end="1942" data-start="1876"><code>text.length == 4 * replacements.length - 1</code></li>
-	<li data-end="2052" data-start="1943">Every placeholder in the <code>text</code> or in any replacement value corresponds to a key in the <code>replacements</code> mapping.</li>
-	<li data-end="2265" data-start="2205">There are no cyclic dependencies between replacement keys.</li>
+	<li data-end="2052" data-start="1943"><code>text</code>&nbsp;或任何替换值中的每个占位符对应&nbsp;<code>replacements</code> 映射中的一个键。</li>
+	<li data-end="2265" data-start="2205">替换键之间没有循环依赖。</li>
 </ul>
 
 <!-- description:end -->