feat: add solutions to lc problems: No.2341~2344

* No.2341.Maximum Number of Pairs in Array
* No.2342.Max Sum of a Pair With Equal Sum of Digits
* No.2343.Query Kth Smallest Trimmed Number
* No.2344.Minimum Deletions to Make Array Divisible

Author: yanglbme

solution/2300-2399/2341.Maximum Number of Pairs in Array/README.md

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+# [2341. 数组能形成多少数对](https://leetcode.cn/problems/maximum-number-of-pairs-in-array)
+
+[English Version](/solution/2300-2399/2341.Maximum%20Number%20of%20Pairs%20in%20Array/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>给你一个下标从 <strong>0</strong> 开始的整数数组 <code>nums</code> 。在一步操作中，你可以执行以下步骤：</p>
+
+<ul>
+	<li>从 <code>nums</code> 选出 <strong>两个</strong> <strong>相等的</strong> 整数</li>
+	<li>从 <code>nums</code> 中移除这两个整数，形成一个 <strong>数对</strong></li>
+</ul>
+
+<p>请你在 <code>nums</code> 上多次执行此操作直到无法继续执行。</p>
+
+<p>返回一个下标从 <strong>0</strong> 开始、长度为 <code>2</code> 的整数数组 <code>answer</code> 作为答案，其中<em> </em><code>answer[0]</code><em> </em>是形成的数对数目，<code>answer[1]</code> 是对 <code>nums</code> 尽可能执行上述操作后剩下的整数数目。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre><strong>输入：</strong>nums = [1,3,2,1,3,2,2]
+<strong>输出：</strong>[3,1]
+<strong>解释：</strong>
+nums[0] 和 nums[3] 形成一个数对，并从 nums 中移除，nums = [3,2,3,2,2] 。
+nums[0] 和 nums[2] 形成一个数对，并从 nums 中移除，nums = [2,2,2] 。
+nums[0] 和 nums[1] 形成一个数对，并从 nums 中移除，nums = [2] 。
+无法形成更多数对。总共形成 3 个数对，nums 中剩下 1 个数字。</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre><strong>输入：</strong>nums = [1,1]
+<strong>输出：</strong>[1,0]
+<strong>解释：</strong>nums[0] 和 nums[1] 形成一个数对，并从 nums 中移除，nums = [] 。
+无法形成更多数对。总共形成 1 个数对，nums 中剩下 0 个数字。</pre>
+
+<p><strong>示例 3：</strong></p>
+
+<pre><strong>输入：</strong>nums = [0]
+<strong>输出：</strong>[0,1]
+<strong>解释：</strong>无法形成数对，nums 中剩下 1 个数字。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 100</code></li>
+	<li><code>0 &lt;= nums[i] &lt;= 100</code></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
+class Solution:
+    def numberOfPairs(self, nums: List[int]) -> List[int]:
+        cnt = Counter(nums)
+        s = sum(v // 2 for v in cnt.values())
+        return [s, len(nums) - s * 2]
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+class Solution {
+    public int[] numberOfPairs(int[] nums) {
+        int[] cnt = new int[101];
+        for (int v : nums) {
+            ++cnt[v];
+        }
+        int s = 0;
+        for (int v : cnt) {
+            s += v / 2;
+        }
+        return new int[]{s, nums.length - s * 2};
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    vector<int> numberOfPairs(vector<int>& nums) {
+        vector<int> cnt(101);
+        for (int v : nums) ++cnt[v];
+        int s = 0;
+        for (int v : cnt) s += v / 2;
+        vector<int> ans(2);
+        ans[0] = s;
+        ans[1] = nums.size() - s * 2;
+        return ans;
+    }
+};
+```
+
+### **Go**
+
+```go
+func numberOfPairs(nums []int) []int {
+    cnt := make([]int, 101)
+    for _, v := range nums {
+        cnt[v]++
+    }
+    s := 0
+    for _, v := range cnt {
+        s += v / 2
+    }
+    return []int{s, len(nums) - s * 2}
+}
+```
+
+### **TypeScript**
+
+```ts
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2300-2399/2341.Maximum Number of Pairs in Array/README_EN.md

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+# [2341. Maximum Number of Pairs in Array](https://leetcode.com/problems/maximum-number-of-pairs-in-array)
+
+[中文文档](/solution/2300-2399/2341.Maximum%20Number%20of%20Pairs%20in%20Array/README.md)
+
+## Description
+
+<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code>. In one operation, you may do the following:</p>
+
+<ul>
+	<li>Choose <strong>two</strong> integers in <code>nums</code> that are <strong>equal</strong>.</li>
+	<li>Remove both integers from <code>nums</code>, forming a <strong>pair</strong>.</li>
+</ul>
+
+<p>The operation is done on <code>nums</code> as many times as possible.</p>
+
+<p>Return <em>a <strong>0-indexed</strong> integer array </em><code>answer</code><em> of size </em><code>2</code><em> where </em><code>answer[0]</code><em> is the number of pairs that are formed and </em><code>answer[1]</code><em> is the number of leftover integers in </em><code>nums</code><em> after doing the operation as many times as possible</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,3,2,1,3,2,2]
+<strong>Output:</strong> [3,1]
+<strong>Explanation:</strong>
+Form a pair with nums[0] and nums[3] and remove them from nums. Now, nums = [3,2,3,2,2].
+Form a pair with nums[0] and nums[2] and remove them from nums. Now, nums = [2,2,2].
+Form a pair with nums[0] and nums[1] and remove them from nums. Now, nums = [2].
+No more pairs can be formed. A total of 3 pairs have been formed, and there is 1 number leftover in nums.
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,1]
+<strong>Output:</strong> [1,0]
+<strong>Explanation:</strong> Form a pair with nums[0] and nums[1] and remove them from nums. Now, nums = [].
+No more pairs can be formed. A total of 1 pair has been formed, and there are 0 numbers leftover in nums.
+</pre>
+
+<p><strong>Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [0]
+<strong>Output:</strong> [0,1]
+<strong>Explanation:</strong> No pairs can be formed, and there is 1 number leftover in nums.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 100</code></li>
+	<li><code>0 &lt;= nums[i] &lt;= 100</code></li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+class Solution:
+    def numberOfPairs(self, nums: List[int]) -> List[int]:
+        cnt = Counter(nums)
+        s = sum(v // 2 for v in cnt.values())
+        return [s, len(nums) - s * 2]
+```
+
+### **Java**
+
+```java
+class Solution {
+    public int[] numberOfPairs(int[] nums) {
+        int[] cnt = new int[101];
+        for (int v : nums) {
+            ++cnt[v];
+        }
+        int s = 0;
+        for (int v : cnt) {
+            s += v / 2;
+        }
+        return new int[]{s, nums.length - s * 2};
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    vector<int> numberOfPairs(vector<int>& nums) {
+        vector<int> cnt(101);
+        for (int v : nums) ++cnt[v];
+        int s = 0;
+        for (int v : cnt) s += v / 2;
+        vector<int> ans(2);
+        ans[0] = s;
+        ans[1] = nums.size() - s * 2;
+        return ans;
+    }
+};
+```
+
+### **Go**
+
+```go
+func numberOfPairs(nums []int) []int {
+    cnt := make([]int, 101)
+    for _, v := range nums {
+        cnt[v]++
+    }
+    s := 0
+    for _, v := range cnt {
+        s += v / 2
+    }
+    return []int{s, len(nums) - s * 2}
+}
+```
+
+### **TypeScript**
+
+```ts
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2300-2399/2341.Maximum Number of Pairs in Array/Solution.cpp

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+class Solution {
+public:
+    vector<int> numberOfPairs(vector<int>& nums) {
+        vector<int> cnt(101);
+        for (int v : nums) ++cnt[v];
+        int s = 0;
+        for (int v : cnt) s += v / 2;
+        vector<int> ans(2);
+        ans[0] = s;
+        ans[1] = nums.size() - s * 2;
+        return ans;
+    }
+};

solution/2300-2399/2341.Maximum Number of Pairs in Array/Solution.go

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+func numberOfPairs(nums []int) []int {
+    cnt := make([]int, 101)
+    for _, v := range nums {
+        cnt[v]++
+    }
+    s := 0
+    for _, v := range cnt {
+        s += v / 2
+    }
+    return []int{s, len(nums) - s * 2}
+}

solution/2300-2399/2341.Maximum Number of Pairs in Array/Solution.java

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+class Solution {
+    public int[] numberOfPairs(int[] nums) {
+        int[] cnt = new int[101];
+        for (int v : nums) {
+            ++cnt[v];
+        }
+        int s = 0;
+        for (int v : cnt) {
+            s += v / 2;
+        }
+        return new int[]{s, nums.length - s * 2};
+    }
+}

solution/2300-2399/2341.Maximum Number of Pairs in Array/Solution.py

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+class Solution:
+    def numberOfPairs(self, nums: List[int]) -> List[int]:
+        cnt = Counter(nums)
+        s = sum(v // 2 for v in cnt.values())
+        return [s, len(nums) - s * 2]