feat: add solutions to lc problems: No.2335~2338

* No.2335.Minimum Amount of Time to Fill Cups
* No.2336.Smallest Number in Infinite Set
* No.2337.Move Pieces to Obtain a String
* No.2338.Count the Number of Ideal Arrays

Author: yanglbme

solution/0500-0599/0561.Array Partition I/Solution.java

@@ -1,26 +1,26 @@
-# [561. 数组拆分 I](https://leetcode.cn/problems/array-partition-i)
+# [561. 数组拆分](https://leetcode.cn/problems/array-partition)
 
-[English Version](/solution/0500-0599/0561.Array%20Partition%20I/README_EN.md)
+[English Version](/solution/0500-0599/0561.Array%20Partition/README_EN.md)
 
 ## 题目描述
 
 <!-- 这里写题目描述 -->
 
-<p>给定长度为 <code>2n</code><strong> </strong>的整数数组 <code>nums</code> ，你的任务是将这些数分成 <code>n</code><strong> </strong>对, 例如 <code>(a<sub>1</sub>, b<sub>1</sub>), (a<sub>2</sub>, b<sub>2</sub>), ..., (a<sub>n</sub>, b<sub>n</sub>)</code> ，使得从 <code>1</code> 到 <code>n</code> 的 <code>min(a<sub>i</sub>, b<sub>i</sub>)</code> 总和最大。</p>
+<p>给定长度为&nbsp;<code>2n</code><strong>&nbsp;</strong>的整数数组 <code>nums</code> ，你的任务是将这些数分成&nbsp;<code>n</code><strong> </strong>对, 例如 <code>(a<sub>1</sub>, b<sub>1</sub>), (a<sub>2</sub>, b<sub>2</sub>), ..., (a<sub>n</sub>, b<sub>n</sub>)</code> ，使得从 <code>1</code> 到&nbsp;<code>n</code> 的 <code>min(a<sub>i</sub>, b<sub>i</sub>)</code> 总和最大。</p>
 
 <p>返回该 <strong>最大总和</strong> 。</p>
 
-<p> </p>
+<p>&nbsp;</p>
 
 <p><strong>示例 1：</strong></p>
 
 <pre>
 <strong>输入：</strong>nums = [1,4,3,2]
 <strong>输出：</strong>4
 <strong>解释：</strong>所有可能的分法（忽略元素顺序）为：
-1. (1, 4), (2, 3) -> min(1, 4) + min(2, 3) = 1 + 2 = 3
-2. (1, 3), (2, 4) -> min(1, 3) + min(2, 4) = 1 + 2 = 3
-3. (1, 2), (3, 4) -> min(1, 2) + min(3, 4) = 1 + 3 = 4
+1. (1, 4), (2, 3) -&gt; min(1, 4) + min(2, 3) = 1 + 2 = 3
+2. (1, 3), (2, 4) -&gt; min(1, 3) + min(2, 4) = 1 + 2 = 3
+3. (1, 2), (3, 4) -&gt; min(1, 2) + min(3, 4) = 1 + 3 = 4
 所以最大总和为 4</pre>
 
 <p><strong>示例 2：</strong></p>
@@ -31,20 +31,23 @@
 <strong>解释：</strong>最优的分法为 (2, 1), (2, 5), (6, 6). min(2, 1) + min(2, 5) + min(6, 6) = 1 + 2 + 6 = 9
 </pre>
 
-<p> </p>
+<p>&nbsp;</p>
 
 <p><strong>提示：</strong></p>
 
 <ul>
-	<li><code>1 <= n <= 10<sup>4</sup></code></li>
+	<li><code>1 &lt;= n &lt;= 10<sup>4</sup></code></li>
 	<li><code>nums.length == 2 * n</code></li>
-	<li><code>-10<sup>4</sup> <= nums[i] <= 10<sup>4</sup></code></li>
+	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
 </ul>
 
+
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：排序**
+
 先排序，然后求相邻的两个元素的最小值，得到的总和即为结果。
 
 <!-- tabs:start -->
@@ -67,15 +70,42 @@ class Solution:
 class Solution {
     public int arrayPairSum(int[] nums) {
         Arrays.sort(nums);
-        int res = 0;
-        for (int i = 0, n = nums.length; i < n; i += 2) {
-            res += nums[i];
+        int ans = 0;
+        for (int i = 0; i < nums.length; i += 2) {
+            ans += nums[i];
         }
-        return res;
+        return ans;
     }
 }
 ```
 
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int arrayPairSum(vector<int>& nums) {
+        sort(nums.begin(), nums.end());
+        int ans = 0;
+        for (int i = 0; i < nums.size(); i += 2) ans += nums[i];
+        return ans;
+    }
+};
+```
+
+### **Go**
+
+```go
+func arrayPairSum(nums []int) int {
+	sort.Ints(nums)
+	ans := 0
+	for i := 0; i < len(nums); i += 2 {
+		ans += nums[i]
+	}
+	return ans
+}
+```
+
 ### **JavaScript**
 
 ```js
@@ -85,11 +115,11 @@ class Solution {
  */
 var arrayPairSum = function (nums) {
     nums.sort((a, b) => a - b);
-    let res = 0;
-    for (let i = 0, n = nums.length; i < n; i += 2) {
-        res += nums[i];
+    let ans = 0;
+    for (let i = 0; i < nums.length; i += 2) {
+        ans += nums[i];
     }
-    return res;
+    return ans;
 };
 ```
 
@@ -111,6 +141,12 @@ impl Solution {
 }
 ```
 
+### **TypeScript**
+
+```ts
+
+```
+
 ### **...**
 
 ```

solution/0500-0599/0561.Array Partition I/README.md solution/0500-0599/0561.Array Partition/README.md

@@ -1,6 +1,6 @@
-# [561. Array Partition I](https://leetcode.com/problems/array-partition-i)
+# [561. Array Partition](https://leetcode.com/problems/array-partition)
 
-[中文文档](/solution/0500-0599/0561.Array%20Partition%20I/README.md)
+[中文文档](/solution/0500-0599/0561.Array%20Partition/README.md)
 
 ## Description
 
@@ -35,6 +35,7 @@ So the maximum possible sum is 4.</pre>
 	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
 </ul>
 
+
 ## Solutions
 
 <!-- tabs:start -->
@@ -53,12 +54,39 @@ class Solution:
 class Solution {
     public int arrayPairSum(int[] nums) {
         Arrays.sort(nums);
-        int res = 0;
-        for (int i = 0, n = nums.length; i < n; i += 2) {
-            res += nums[i];
+        int ans = 0;
+        for (int i = 0; i < nums.length; i += 2) {
+            ans += nums[i];
         }
-        return res;
+        return ans;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int arrayPairSum(vector<int>& nums) {
+        sort(nums.begin(), nums.end());
+        int ans = 0;
+        for (int i = 0; i < nums.size(); i += 2) ans += nums[i];
+        return ans;
     }
+};
+```
+
+### **Go**
+
+```go
+func arrayPairSum(nums []int) int {
+	sort.Ints(nums)
+	ans := 0
+	for i := 0; i < len(nums); i += 2 {
+		ans += nums[i]
+	}
+	return ans
 }
 ```
 
@@ -71,11 +99,11 @@ class Solution {
  */
 var arrayPairSum = function (nums) {
     nums.sort((a, b) => a - b);
-    let res = 0;
-    for (let i = 0, n = nums.length; i < n; i += 2) {
-        res += nums[i];
+    let ans = 0;
+    for (let i = 0; i < nums.length; i += 2) {
+        ans += nums[i];
     }
-    return res;
+    return ans;
 };
 ```
 
@@ -97,6 +125,12 @@ impl Solution {
 }
 ```
 
+### **TypeScript**
+
+```ts
+
+```
+
 ### **...**
 
 ```

solution/0500-0599/0561.Array Partition I/README_EN.md solution/0500-0599/0561.Array Partition/README_EN.md

@@ -0,0 +1,9 @@
+class Solution {
+public:
+    int arrayPairSum(vector<int>& nums) {
+        sort(nums.begin(), nums.end());
+        int ans = 0;
+        for (int i = 0; i < nums.size(); i += 2) ans += nums[i];
+        return ans;
+    }
+};

solution/0500-0599/0561.Array Partition/Solution.cpp

@@ -0,0 +1,8 @@
+func arrayPairSum(nums []int) int {
+	sort.Ints(nums)
+	ans := 0
+	for i := 0; i < len(nums); i += 2 {
+		ans += nums[i]
+	}
+	return ans
+}

solution/0500-0599/0561.Array Partition/Solution.go

@@ -0,0 +1,10 @@
+class Solution {
+    public int arrayPairSum(int[] nums) {
+        Arrays.sort(nums);
+        int ans = 0;
+        for (int i = 0; i < nums.length; i += 2) {
+            ans += nums[i];
+        }
+        return ans;
+    }
+}

solution/0500-0599/0561.Array Partition/Solution.java

@@ -4,9 +4,9 @@
  */
 var arrayPairSum = function (nums) {
     nums.sort((a, b) => a - b);
-    let res = 0;
-    for (let i = 0, n = nums.length; i < n; i += 2) {
-        res += nums[i];
+    let ans = 0;
+    for (let i = 0; i < nums.length; i += 2) {
+        ans += nums[i];
     }
-    return res;
+    return ans;
 };

solution/0500-0599/0561.Array Partition I/Solution.js solution/0500-0599/0561.Array Partition/Solution.js

@@ -0,0 +1,85 @@
+# [2330. Valid Palindrome IV](https://leetcode.cn/problems/valid-palindrome-iv)
+
+[English Version](/solution/2300-2399/2330.Valid%20Palindrome%20IV/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>You are given a <strong>0-indexed</strong> string <code>s</code> consisting of only lowercase English letters. In one operation, you can change <strong>any</strong> character of <code>s</code> to any <strong>other</strong> character.</p>
+
+<p>Return <code>true</code><em> if you can make </em><code>s</code><em> a palindrome after performing <strong>exactly</strong> one or two operations, or return </em><code>false</code><em> otherwise.</em></p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;abcdba&quot;
+<strong>Output:</strong> true
+<strong>Explanation:</strong> One way to make s a palindrome using 1 operation is:
+- Change s[2] to &#39;d&#39;. Now, s = &quot;abddba&quot;.
+One operation could be performed to make s a palindrome so return true.
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;aa&quot;
+<strong>Output:</strong> true
+<strong>Explanation:</strong> One way to make s a palindrome using 2 operations is:
+- Change s[0] to &#39;b&#39;. Now, s = &quot;ba&quot;.
+- Change s[1] to &#39;b&#39;. Now, s = &quot;bb&quot;.
+Two operations could be performed to make s a palindrome so return true.
+</pre>
+
+<p><strong>Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;abcdef&quot;
+<strong>Output:</strong> false
+<strong>Explanation:</strong> It is not possible to make s a palindrome using one or two operations so return false.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>s</code> consists only of lowercase English letters.</li>
+</ul>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+<!-- tabs:start -->
+
+### **Python3**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+
+```
+
+### **TypeScript**
+
+```ts
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->