feat: add solutions to lc problems: No.2287~2290

* No.2287.Rearrange Characters to Make Target String
* No.2288.Apply Discount to Prices
* No.2289.Steps to Make Array Non-decreasing
* No.2290.Minimum Obstacle Removal to Reach Corner

Author: yanglbme

README.md

@@ -86,6 +86,7 @@
 -   [访问所有节点的最短路径](/solution/0800-0899/0847.Shortest%20Path%20Visiting%20All%20Nodes/README.md) - `BFS`、`最小步数模型`、`A* 算法`
 -   [为高尔夫比赛砍树](/solution/0600-0699/0675.Cut%20Off%20Trees%20for%20Golf%20Event/README.md) - `BFS`、`A* 算法`
 -   [使网格图至少有一条有效路径的最小代价](/solution/1300-1399/1368.Minimum%20Cost%20to%20Make%20at%20Least%20One%20Valid%20Path%20in%20a%20Grid/README.md) - `双端队列 BFS`
+-   [到达角落需要移除障碍物的最小数目](/solution/2200-2299/2290.Minimum%20Obstacle%20Removal%20to%20Reach%20Corner/README.md) - `双端队列 BFS`
 -   [迷宫](/solution/0400-0499/0490.The%20Maze/README.md) - `DFS`、`连通性模型`、`Flood Fill 算法`
 -   [单词搜索](/solution/0000-0099/0079.Word%20Search/README.md) - `DFS`、`搜索顺序`、`回溯`
 -   [黄金矿工](/solution/1200-1299/1219.Path%20with%20Maximum%20Gold/README.md) - `DFS`、`搜索顺序`、`回溯`

README_EN.md

@@ -84,6 +84,7 @@ Complete solutions to [LeetCode](https://leetcode.com/problemset/all/), [LCOF](h
 -   [Shortest Path Visiting All Nodes](/solution/0800-0899/0847.Shortest%20Path%20Visiting%20All%20Nodes/README_EN.md) - `BFS`, `Minimum steps model`, `A* search`
 -   [Cut Off Trees for Golf Event](/solution/0600-0699/0675.Cut%20Off%20Trees%20for%20Golf%20Event/README_EN.md) - `BFS`, `A* search`
 -   [Minimum Cost to Make at Least One Valid Path in a Grid](/solution/1300-1399/1368.Minimum%20Cost%20to%20Make%20at%20Least%20One%20Valid%20Path%20in%20a%20Grid/README_EN.md) - `BFS using deque`
+-   [Minimum Cost to Make at Least One Valid Path in a Grid](/solution/2200-2299/2290.Minimum%20Obstacle%20Removal%20to%20Reach%20Corner/README_EN.md) - `BFS using deque`
 -   [The Maze](/solution/0400-0499/0490.The%20Maze/README_EN.md) - `DFS, Flood fill`
 -   [Word Search](/solution/0000-0099/0079.Word%20Search/README_EN.md) - `DFS`, `Backtracking`
 -   [Path with Maximum Gold](/solution/1200-1299/1219.Path%20with%20Maximum%20Gold/README_EN.md) - `DFS`, `Backtracking`

solution/1300-1399/1368.Minimum Cost to Make at Least One Valid Path in a Grid/README.md

@@ -79,7 +79,9 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
-双端队列 BFS。本题实际上也是最短路模型，只不过求解的是改变方向的最小次数。
+**方法一：双端队列 BFS**
+
+本题实际上也是最短路模型，只不过求解的是改变方向的最小次数。
 
 在一个边权只有 0、1 的无向图中搜索最短路径可以使用双端队列进行 BFS。其原理是当前可以扩展到的点的权重为 0 时，将其加入队首；权重为 1 时，将其加入队尾。
 

solution/2200-2299/2282.Number of People That Can Be Seen in a Grid/README.md

@@ -0,0 +1,93 @@
+# [2282. Number of People That Can Be Seen in a Grid](https://leetcode.cn/problems/number-of-people-that-can-be-seen-in-a-grid)
+
+[English Version](/solution/2200-2299/2282.Number%20of%20People%20That%20Can%20Be%20Seen%20in%20a%20Grid/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>You are given an <code>m x n</code> <strong>0-indexed</strong> 2D array of positive integers <code>heights</code> where <code>heights[i][j]</code> is the height of the person standing at position <code>(i, j)</code>.</p>
+
+<p>A person standing at position <code>(row<sub>1</sub>, col<sub>1</sub>)</code> can see a person standing at position <code>(row<sub>2</sub>, col<sub>2</sub>)</code> if:</p>
+
+<ul>
+	<li>The person at <code>(row<sub>2</sub>, col<sub>2</sub>)</code> is to the right <strong>or</strong> below the person at <code>(row<sub>1</sub>, col<sub>1</sub>)</code>. More formally, this means that either <code>row<sub>1</sub> == row<sub>2</sub></code> and <code>col<sub>1</sub> &lt; col<sub>2</sub></code> <strong>or</strong> <code>row<sub>1</sub> &lt; row<sub>2</sub></code> and <code>col<sub>1</sub> == col<sub>2</sub></code>.</li>
+	<li>Everyone in between them is shorter than <strong>both</strong> of them.</li>
+</ul>
+
+<p>Return<em> an </em><code>m x n</code><em> 2D array of integers </em><code>answer</code><em> where </em><code>answer[i][j]</code><em> is the number of people that the person at position </em><code>(i, j)</code><em> can see.</em></p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+<img src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2200-2299/2282.Number%20of%20People%20That%20Can%20Be%20Seen%20in%20a%20Grid/images/image-20220524180458-1.png" style="width: 700px; height: 164px;" />
+<pre>
+<strong>Input:</strong> heights = [[3,1,4,2,5]]
+<strong>Output:</strong> [[2,1,2,1,0]]
+<strong>Explanation:</strong>
+- The person at (0, 0) can see the people at (0, 1) and (0, 2).
+  Note that he cannot see the person at (0, 4) because the person at (0, 2) is taller than him.
+- The person at (0, 1) can see the person at (0, 2).
+- The person at (0, 2) can see the people at (0, 3) and (0, 4).
+- The person at (0, 3) can see the person at (0, 4).
+- The person at (0, 4) cannot see anybody.
+</pre>
+
+<p><strong>Example 2:</strong></p>
+<img src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2200-2299/2282.Number%20of%20People%20That%20Can%20Be%20Seen%20in%20a%20Grid/images/image-20220523113533-2.png" style="width: 400px; height: 249px;" />
+<pre>
+<strong>Input:</strong> heights = [[5,1],[3,1],[4,1]]
+<strong>Output:</strong> [[3,1],[2,1],[1,0]]
+<strong>Explanation:</strong>
+- The person at (0, 0) can see the people at (0, 1), (1, 0) and (2, 0).
+- The person at (0, 1) can see the person at (1, 1).
+- The person at (1, 0) can see the people at (1, 1) and (2, 0).
+- The person at (1, 1) can see the person at (2, 1).
+- The person at (2, 0) can see the person at (2, 1).
+- The person at (2, 1) cannot see anybody.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= heights.length &lt;= 400</code></li>
+	<li><code>1 &lt;= heights[i].length &lt;= 400</code></li>
+	<li><code>1 &lt;= heights[i][j] &lt;= 10<sup>5</sup></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
+
+```
+
+### **TypeScript**
+
+```ts
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2200-2299/2282.Number of People That Can Be Seen in a Grid/README_EN.md

@@ -0,0 +1,84 @@
+# [2282. Number of People That Can Be Seen in a Grid](https://leetcode.com/problems/number-of-people-that-can-be-seen-in-a-grid)
+
+[中文文档](/solution/2200-2299/2282.Number%20of%20People%20That%20Can%20Be%20Seen%20in%20a%20Grid/README.md)
+
+## Description
+
+<p>You are given an <code>m x n</code> <strong>0-indexed</strong> 2D array of positive integers <code>heights</code> where <code>heights[i][j]</code> is the height of the person standing at position <code>(i, j)</code>.</p>
+
+<p>A person standing at position <code>(row<sub>1</sub>, col<sub>1</sub>)</code> can see a person standing at position <code>(row<sub>2</sub>, col<sub>2</sub>)</code> if:</p>
+
+<ul>
+	<li>The person at <code>(row<sub>2</sub>, col<sub>2</sub>)</code> is to the right <strong>or</strong> below the person at <code>(row<sub>1</sub>, col<sub>1</sub>)</code>. More formally, this means that either <code>row<sub>1</sub> == row<sub>2</sub></code> and <code>col<sub>1</sub> &lt; col<sub>2</sub></code> <strong>or</strong> <code>row<sub>1</sub> &lt; row<sub>2</sub></code> and <code>col<sub>1</sub> == col<sub>2</sub></code>.</li>
+	<li>Everyone in between them is shorter than <strong>both</strong> of them.</li>
+</ul>
+
+<p>Return<em> an </em><code>m x n</code><em> 2D array of integers </em><code>answer</code><em> where </em><code>answer[i][j]</code><em> is the number of people that the person at position </em><code>(i, j)</code><em> can see.</em></p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+<img src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2200-2299/2282.Number%20of%20People%20That%20Can%20Be%20Seen%20in%20a%20Grid/images/image-20220524180458-1.png" style="width: 700px; height: 164px;" />
+<pre>
+<strong>Input:</strong> heights = [[3,1,4,2,5]]
+<strong>Output:</strong> [[2,1,2,1,0]]
+<strong>Explanation:</strong>
+- The person at (0, 0) can see the people at (0, 1) and (0, 2).
+  Note that he cannot see the person at (0, 4) because the person at (0, 2) is taller than him.
+- The person at (0, 1) can see the person at (0, 2).
+- The person at (0, 2) can see the people at (0, 3) and (0, 4).
+- The person at (0, 3) can see the person at (0, 4).
+- The person at (0, 4) cannot see anybody.
+</pre>
+
+<p><strong>Example 2:</strong></p>
+<img src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2200-2299/2282.Number%20of%20People%20That%20Can%20Be%20Seen%20in%20a%20Grid/images/image-20220523113533-2.png" style="width: 400px; height: 249px;" />
+<pre>
+<strong>Input:</strong> heights = [[5,1],[3,1],[4,1]]
+<strong>Output:</strong> [[3,1],[2,1],[1,0]]
+<strong>Explanation:</strong>
+- The person at (0, 0) can see the people at (0, 1), (1, 0) and (2, 0).
+- The person at (0, 1) can see the person at (1, 1).
+- The person at (1, 0) can see the people at (1, 1) and (2, 0).
+- The person at (1, 1) can see the person at (2, 1).
+- The person at (2, 0) can see the person at (2, 1).
+- The person at (2, 1) cannot see anybody.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= heights.length &lt;= 400</code></li>
+	<li><code>1 &lt;= heights[i].length &lt;= 400</code></li>
+	<li><code>1 &lt;= heights[i][j] &lt;= 10<sup>5</sup></code></li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+
+```
+
+### **Java**
+
+```java
+
+```
+
+### **TypeScript**
+
+```ts
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2200-2299/2282.Number of People That Can Be Seen in a Grid/images/image-20220523113533-2.png

@@ -0,0 +1,83 @@
+# [2283. 判断一个数的数字计数是否等于数位的值](https://leetcode.cn/problems/check-if-number-has-equal-digit-count-and-digit-value)
+
+[English Version](/solution/2200-2299/2283.Check%20if%20Number%20Has%20Equal%20Digit%20Count%20and%20Digit%20Value/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>&nbsp;开始长度为 <code>n</code>&nbsp;的字符串&nbsp;<code>num</code>&nbsp;，它只包含数字。</p>
+
+<p>如果对于 <strong>每个</strong><em>&nbsp;</em><code>0 &lt;= i &lt; n</code>&nbsp;的下标&nbsp;<code>i</code>&nbsp;，都满足数位<em>&nbsp;</em><code>i</code>&nbsp;在 <code>num</code>&nbsp;中出现了&nbsp;<code>num[i]</code>次，那么请你返回&nbsp;<code>true</code>&nbsp;，否则返回&nbsp;<code>false</code>&nbsp;。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre><b>输入：</b>num = "1210"
+<b>输出：</b>true
+<strong>解释：</strong>
+num[0] = '1' 。数字 0 在 num 中出现了一次。
+num[1] = '2' 。数字 1 在 num 中出现了两次。
+num[2] = '1' 。数字 2 在 num 中出现了一次。
+num[3] = '0' 。数字 3 在 num 中出现了零次。
+"1210" 满足题目要求条件，所以返回 true 。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre><b>输入：</b>num = "030"
+<b>输出：</b>false
+<strong>解释：</strong>
+num[0] = '0' 。数字 0 应该出现 0 次，但是在 num 中出现了一次。
+num[1] = '3' 。数字 1 应该出现 3 次，但是在 num 中出现了零次。
+num[2] = '0' 。数字 2 在 num 中出现了 0 次。
+下标 0 和 1 都违反了题目要求，所以返回 false 。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>n == num.length</code></li>
+	<li><code>1 &lt;= n &lt;= 10</code></li>
+	<li><code>num</code>&nbsp;只包含数字。</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 -->

solution/2200-2299/2282.Number of People That Can Be Seen in a Grid/images/image-20220524180458-1.png

@@ -0,0 +1,75 @@
+# [2283. Check if Number Has Equal Digit Count and Digit Value](https://leetcode.com/problems/check-if-number-has-equal-digit-count-and-digit-value)
+
+[中文文档](/solution/2200-2299/2283.Check%20if%20Number%20Has%20Equal%20Digit%20Count%20and%20Digit%20Value/README.md)
+
+## Description
+
+<p>You are given a <strong>0-indexed</strong> string <code>num</code> of length <code>n</code> consisting of digits.</p>
+
+<p>Return <code>true</code> <em>if for <strong>every</strong> index </em><code>i</code><em> in the range </em><code>0 &lt;= i &lt; n</code><em>, the digit </em><code>i</code><em> occurs </em><code>num[i]</code><em> times in </em><code>num</code><em>, otherwise return </em><code>false</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> num = &quot;1210&quot;
+<strong>Output:</strong> true
+<strong>Explanation:</strong>
+num[0] = &#39;1&#39;. The digit 0 occurs once in num.
+num[1] = &#39;2&#39;. The digit 1 occurs twice in num.
+num[2] = &#39;1&#39;. The digit 2 occurs once in num.
+num[3] = &#39;0&#39;. The digit 3 occurs zero times in num.
+The condition holds true for every index in &quot;1210&quot;, so return true.
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> num = &quot;030&quot;
+<strong>Output:</strong> false
+<strong>Explanation:</strong>
+num[0] = &#39;0&#39;. The digit 0 should occur zero times, but actually occurs twice in num.
+num[1] = &#39;3&#39;. The digit 1 should occur three times, but actually occurs zero times in num.
+num[2] = &#39;0&#39;. The digit 2 occurs zero times in num.
+The indices 0 and 1 both violate the condition, so return false.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>n == num.length</code></li>
+	<li><code>1 &lt;= n &lt;= 10</code></li>
+	<li><code>num</code> consists of digits.</li>
+</ul>
+
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+
+```
+
+### **Java**
+
+```java
+
+```
+
+### **TypeScript**
+
+```ts
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->