feat: add new leetcode problems: No.1850+

Author: yanglbme

solution/1800-1899/1852.Distinct Numbers in Each Subarray/README.md

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+# [1852. ](https://leetcode-cn.com/problems/distinct-numbers-in-each-subarray)
+
+[English Version](/solution/1800-1899/1852.Distinct%20Numbers%20in%20Each%20Subarray/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+None
+
+## 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
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/1800-1899/1852.Distinct Numbers in Each Subarray/README_EN.md

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+# [1852. Distinct Numbers in Each Subarray](https://leetcode.com/problems/distinct-numbers-in-each-subarray)
+
+[中文文档](/solution/1800-1899/1852.Distinct%20Numbers%20in%20Each%20Subarray/README.md)
+
+## Description
+
+<p>Given an integer array <code>nums</code> and an integer <code>k</code>, you are asked to construct the array <code>ans</code> of size <code>n-k+1</code> where <code>ans[i]</code> is the number of <strong>distinct</strong> numbers in the subarray <code>nums[i:i+k-1] = [nums[i], nums[i+1], ..., nums[i+k-1]]</code>.</p>
+
+
+
+<p>Return <em>the array </em><code>ans</code>.</p>
+
+
+
+<p>&nbsp;</p>
+
+<p><strong>Example 1:</strong></p>
+
+
+
+<pre>
+
+<strong>Input:</strong> nums = [1,2,3,2,2,1,3], k = 3
+
+<strong>Output:</strong> [3,2,2,2,3]
+
+<strong>Explanation: </strong>The number of distinct elements in each subarray goes as follows:
+
+- nums[0:2] = [1,2,3] so ans[0] = 3
+
+- nums[1:3] = [2,3,2] so ans[1] = 2
+
+- nums[2:4] = [3,2,2] so ans[2] = 2
+
+- nums[3:5] = [2,2,1] so ans[3] = 2
+
+- nums[4:6] = [2,1,3] so ans[4] = 3
+
+</pre>
+
+
+
+<p><strong>Example 2:</strong></p>
+
+
+
+<pre>
+
+<strong>Input:</strong> nums = [1,1,1,1,2,3,4], k = 4
+
+<strong>Output:</strong> [1,2,3,4]
+
+<strong>Explanation: </strong>The number of distinct elements in each subarray goes as follows:
+
+- nums[0:3] = [1,1,1,1] so ans[0] = 1
+
+- nums[1:4] = [1,1,1,2] so ans[1] = 2
+
+- nums[2:5] = [1,1,2,3] so ans[2] = 3
+
+- nums[3:6] = [1,2,3,4] so ans[3] = 4
+
+</pre>
+
+
+
+<p>&nbsp;</p>
+
+<p><strong>Constraints:</strong></p>
+
+
+
+<ul>
+	<li><code>1 &lt;= k &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+
+```
+
+### **Java**
+
+```java
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/1800-1899/1853.Convert Date Format/README.md

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+# [1853. ](https://leetcode-cn.com/problems/convert-date-format)
+
+[English Version](/solution/1800-1899/1853.Convert%20Date%20Format/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+None
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+<!-- tabs:start -->
+
+### **SQL**
+
+```sql
+
+```
+
+<!-- tabs:end -->

solution/1800-1899/1853.Convert Date Format/README_EN.md

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+# [1853. Convert Date Format](https://leetcode.com/problems/convert-date-format)
+
+[中文文档](/solution/1800-1899/1853.Convert%20Date%20Format/README.md)
+
+## Description
+
+<p>Table: <code>Days</code></p>
+
+<pre>
++-------------+------+
+| Column Name | Type |
++-------------+------+
+| day         | date |
++-------------+------+
+day is the primary key for this table.
+</pre>
+
+<p>&nbsp;</p>
+
+<p>Write an SQL query to convert each date in <code>Days</code> into a string formatted as <code>&quot;day_name, month_name day, year&quot;</code>.</p>
+
+<p>Return the result table <strong>in any order</strong>.</p>
+
+<p>The query result format is in the following example:</p>
+
+<p>&nbsp;</p>
+
+<pre>
+Days table:
++------------+
+| day        |
++------------+
+| 2022-04-12 |
+| 2021-08-09 |
+| 2020-06-26 |
++------------+
+
+Result table:
++-------------------------+
+| day                     |
++-------------------------+
+| Tuesday, April 12, 2022 |
+| Monday, August 9, 2021  |
+| Friday, June 26, 2020   |
++-------------------------+
+Please note that the output is case-sensitive.
+</pre>
+
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **SQL**
+
+```sql
+
+```
+
+<!-- tabs:end -->

solution/1800-1899/1854.Maximum Population Year/README.md

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+# [1854. 人口最多的年份](https://leetcode-cn.com/problems/maximum-population-year)
+
+[English Version](/solution/1800-1899/1854.Maximum%20Population%20Year/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>给你一个二维整数数组 <code>logs</code> ，其中每个 <code>logs[i] = [birth<sub>i</sub>, death<sub>i</sub>]</code> 表示第 <code>i</code> 个人的出生和死亡年份。</p>
+
+<p>年份 <code>x</code> 的 <strong>人口</strong> 定义为这一年期间活着的人的数目。第 <code>i</code> 个人被计入年份 <code>x</code> 的人口需要满足：<code>x</code> 在闭区间 <code>[birth<sub>i</sub>, death<sub>i</sub> - 1]</code> 内。注意，人不应当计入他们死亡当年的人口中。</p>
+
+<p>返回 <strong>人口最多</strong> 且 <strong>最早</strong> 的年份。</p>
+
+<p> </p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre><strong>输入：</strong>logs = [[1993,1999],[2000,2010]]
+<strong>输出：</strong>1993
+<strong>解释：</strong>人口最多为 1 ，而 1993 是人口为 1 的最早年份。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre><strong>输入：</strong>logs = [[1950,1961],[1960,1971],[1970,1981]]
+<strong>输出：</strong>1960
+<strong>解释：</strong> 
+人口最多为 2 ，分别出现在 1960 和 1970 。
+其中最早年份是 1960 。</pre>
+
+<p> </p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= logs.length &lt;= 100</code></li>
+	<li><code>1950 &lt;= birth<sub>i</sub> &lt; death<sub>i</sub> &lt;= 2050</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
+
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/1800-1899/1854.Maximum Population Year/README_EN.md

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+# [1854. Maximum Population Year](https://leetcode.com/problems/maximum-population-year)
+
+[中文文档](/solution/1800-1899/1854.Maximum%20Population%20Year/README.md)
+
+## Description
+
+<p>You are given a 2D integer array <code>logs</code> where each <code>logs[i] = [birth<sub>i</sub>, death<sub>i</sub>]</code> indicates the birth and death years of the <code>i<sup>th</sup></code> person.</p>
+
+<p>The <strong>population</strong> of some year <code>x</code> is the number of people alive during that year. The <code>i<sup>th</sup></code> person is counted in year <code>x</code>&#39;s population if <code>x</code> is in the <strong>inclusive</strong> range <code>[birth<sub>i</sub>, death<sub>i</sub> - 1]</code>. Note that the person is <strong>not</strong> counted in the year that they die.</p>
+
+<p>Return <em>the <strong>earliest</strong> year with the <strong>maximum population</strong></em>.</p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> logs = [[1993,1999],[2000,2010]]
+<strong>Output:</strong> 1993
+<strong>Explanation:</strong> The maximum population is 1, and 1993 is the earliest year with this population.
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> logs = [[1950,1961],[1960,1971],[1970,1981]]
+<strong>Output:</strong> 1960
+<strong>Explanation:</strong> 
+The maximum population is 2, and it had happened in years 1960 and 1970.
+The earlier year between them is 1960.</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= logs.length &lt;= 100</code></li>
+	<li><code>1950 &lt;= birth<sub>i</sub> &lt; death<sub>i</sub> &lt;= 2050</code></li>
+</ul>
+
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+
+```
+
+### **Java**
+
+```java
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->