[pull] main from doocs:main

solution/2200-2299/2206.Divide Array Into Equal Pairs/README.md

@@ -144,18 +144,6 @@ func divideArray(nums []int) bool {
 }
 ```
 
-#### TypeScript
-
-```ts
-function divideArray(nums: number[]): boolean {
-    const cnt: Record<number, number> = {};
-    for (const x of nums) {
-        cnt[x] = (cnt[x] || 0) + 1;
-    }
-    return Object.values(cnt).every(x => x % 2 === 0);
-}
-```
-
 #### Rust
 
 ```rust
@@ -172,6 +160,24 @@ impl Solution {
 }
 ```
 
+#### TypeScript
+
+```ts
+function divideArray(nums: number[]): boolean {
+    const cnt = Array(501).fill(0);
+
+    for (const x of nums) {
+        cnt[x]++;
+    }
+
+    for (const x of cnt) {
+        if (x & 1) return false;
+    }
+
+    return true;
+}
+```
+
 #### JavaScript
 
 ```js
@@ -180,11 +186,17 @@ impl Solution {
  * @return {boolean}
  */
 var divideArray = function (nums) {
-    const cnt = {};
+    const cnt = Array(501).fill(0);
+
     for (const x of nums) {
-        cnt[x] = (cnt[x] || 0) + 1;
+        cnt[x]++;
     }
-    return Object.values(cnt).every(x => x % 2 === 0);
+
+    for (const x of cnt) {
+        if (x & 1) return false;
+    }
+
+    return true;
 };
 ```
 

solution/2200-2299/2206.Divide Array Into Equal Pairs/README_EN.md

@@ -38,7 +38,7 @@ tags:
 <pre>
 <strong>Input:</strong> nums = [3,2,3,2,2,2]
 <strong>Output:</strong> true
-<strong>Explanation:</strong> 
+<strong>Explanation:</strong>
 There are 6 elements in nums, so they should be divided into 6 / 2 = 3 pairs.
 If nums is divided into the pairs (2, 2), (3, 3), and (2, 2), it will satisfy all the conditions.
 </pre>
@@ -48,7 +48,7 @@ If nums is divided into the pairs (2, 2), (3, 3), and (2, 2), it will satisfy al
 <pre>
 <strong>Input:</strong> nums = [1,2,3,4]
 <strong>Output:</strong> false
-<strong>Explanation:</strong> 
+<strong>Explanation:</strong>
 There is no way to divide nums into 4 / 2 = 2 pairs such that the pairs satisfy every condition.
 </pre>
 
@@ -142,18 +142,6 @@ func divideArray(nums []int) bool {
 }
 ```
 
-#### TypeScript
-
-```ts
-function divideArray(nums: number[]): boolean {
-    const cnt: Record<number, number> = {};
-    for (const x of nums) {
-        cnt[x] = (cnt[x] || 0) + 1;
-    }
-    return Object.values(cnt).every(x => x % 2 === 0);
-}
-```
-
 #### Rust
 
 ```rust
@@ -170,6 +158,24 @@ impl Solution {
 }
 ```
 
+#### TypeScript
+
+```ts
+function divideArray(nums: number[]): boolean {
+    const cnt = Array(501).fill(0);
+
+    for (const x of nums) {
+        cnt[x]++;
+    }
+
+    for (const x of cnt) {
+        if (x & 1) return false;
+    }
+
+    return true;
+}
+```
+
 #### JavaScript
 
 ```js
@@ -178,11 +184,17 @@ impl Solution {
  * @return {boolean}
  */
 var divideArray = function (nums) {
-    const cnt = {};
+    const cnt = Array(501).fill(0);
+
     for (const x of nums) {
-        cnt[x] = (cnt[x] || 0) + 1;
+        cnt[x]++;
     }
-    return Object.values(cnt).every(x => x % 2 === 0);
+
+    for (const x of cnt) {
+        if (x & 1) return false;
+    }
+
+    return true;
 };
 ```
 

solution/2200-2299/2206.Divide Array Into Equal Pairs/Solution.js

@@ -3,9 +3,15 @@
  * @return {boolean}
  */
 var divideArray = function (nums) {
-    const cnt = {};
+    const cnt = Array(501).fill(0);
+
     for (const x of nums) {
-        cnt[x] = (cnt[x] || 0) + 1;
+        cnt[x]++;
     }
-    return Object.values(cnt).every(x => x % 2 === 0);
+
+    for (const x of cnt) {
+        if (x & 1) return false;
+    }
+
+    return true;
 };

solution/2200-2299/2206.Divide Array Into Equal Pairs/Solution.ts

@@ -1,7 +1,13 @@
 function divideArray(nums: number[]): boolean {
-    const cnt: Record<number, number> = {};
+    const cnt = Array(501).fill(0);
+
     for (const x of nums) {
-        cnt[x] = (cnt[x] || 0) + 1;
+        cnt[x]++;
     }
-    return Object.values(cnt).every(x => x % 2 === 0);
+
+    for (const x of cnt) {
+        if (x & 1) return false;
+    }
+
+    return true;
 }

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>