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solution/1900-1999/1909.Remove One Element to Make the Array Strictly Increasing/README.md

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+# [1909. 删除一个元素使数组严格递增](https://leetcode-cn.com/problems/remove-one-element-to-make-the-array-strictly-increasing)
+
+[English Version](/solution/1900-1999/1909.Remove%20One%20Element%20to%20Make%20the%20Array%20Strictly%20Increasing/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> ，如果 <strong>恰好</strong> 删除 <strong>一个</strong> 元素后，数组 <strong>严格递增</strong> ，那么请你返回 <code>true</code> ，否则返回 <code>false</code> 。如果数组本身已经是严格递增的，请你也返回 <code>true</code> 。</p>
+
+<p>数组 <code>nums</code> 是 <strong>严格递增</strong> 的定义为：对于任意下标的 <code>1 &lt;= i &lt; nums.length</code> 都满足 <code>nums[i - 1] &lt; nums[i]</code> 。</p>
+
+<p> </p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre><b>输入：</b>nums = [1,2,<strong>10</strong>,5,7]
+<b>输出：</b>true
+<b>解释：</b>从 nums 中删除下标 2 处的 10 ，得到 [1,2,5,7] 。
+[1,2,5,7] 是严格递增的，所以返回 true 。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre><b>输入：</b>nums = [2,3,1,2]
+<b>输出：</b>false
+<b>解释：</b>
+[3,1,2] 是删除下标 0 处元素后得到的结果。
+[2,1,2] 是删除下标 1 处元素后得到的结果。
+[2,3,2] 是删除下标 2 处元素后得到的结果。
+[2,3,1] 是删除下标 3 处元素后得到的结果。
+没有任何结果数组是严格递增的，所以返回 false 。</pre>
+
+<p><strong>示例 3：</strong></p>
+
+<pre><b>输入：</b>nums = [1,1,1]
+<b>输出：</b>false
+<b>解释：</b>删除任意元素后的结果都是 [1,1] 。
+[1,1] 不是严格递增的，所以返回 false 。
+</pre>
+
+<p><strong>示例 4：</strong></p>
+
+<pre><b>输入：</b>nums = [1,2,3]
+<b>输出：</b>true
+<b>解释：</b>[1,2,3] 已经是严格递增的，所以返回 true 。
+</pre>
+
+<p> </p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>2 &lt;= nums.length &lt;= 1000</code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 1000</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/1900-1999/1909.Remove One Element to Make the Array Strictly Increasing/README_EN.md

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+# [1909. Remove One Element to Make the Array Strictly Increasing](https://leetcode.com/problems/remove-one-element-to-make-the-array-strictly-increasing)
+
+[中文文档](/solution/1900-1999/1909.Remove%20One%20Element%20to%20Make%20the%20Array%20Strictly%20Increasing/README.md)
+
+## Description
+
+<p>Given a <strong>0-indexed</strong> integer array <code>nums</code>, return <code>true</code> <em>if it can be made <strong>strictly increasing</strong> after removing <strong>exactly one</strong> element, or </em><code>false</code><em> otherwise. If the array is already strictly increasing, return </em><code>true</code>.</p>
+
+<p>The array <code>nums</code> is <strong>strictly increasing</strong> if <code>nums[i - 1] &lt; nums[i]</code> for each index <code>(1 &lt;= i &lt; nums.length).</code></p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,2,<u>10</u>,5,7]
+<strong>Output:</strong> true
+<strong>Explanation:</strong> By removing 10 at index 2 from nums, it becomes [1,2,5,7].
+[1,2,5,7] is strictly increasing, so return true.
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [2,3,1,2]
+<strong>Output:</strong> false
+<strong>Explanation:</strong>
+[3,1,2] is the result of removing the element at index 0.
+[2,1,2] is the result of removing the element at index 1.
+[2,3,2] is the result of removing the element at index 2.
+[2,3,1] is the result of removing the element at index 3.
+No resulting array is strictly increasing, so return false.</pre>
+
+<p><strong>Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,1,1]
+<strong>Output:</strong> false
+<strong>Explanation:</strong> The result of removing any element is [1,1].
+[1,1] is not strictly increasing, so return false.
+</pre>
+
+<p><strong>Example 4:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,2,3]
+<strong>Output:</strong> true
+<strong>Explanation:</strong> [1,2,3] is already strictly increasing, so return true.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>2 &lt;= nums.length &lt;= 1000</code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 1000</code></li>
+</ul>
+
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+
+```
+
+### **Java**
+
+```java
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/1900-1999/1910.Remove All Occurrences of a Substring/README.md

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+# [1910. 删除一个字符串中所有出现的给定子字符串](https://leetcode-cn.com/problems/remove-all-occurrences-of-a-substring)
+
+[English Version](/solution/1900-1999/1910.Remove%20All%20Occurrences%20of%20a%20Substring/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>给你两个字符串 <code>s</code> 和 <code>part</code> ，请你对 <code>s</code> 反复执行以下操作直到 <b>所有</b> 子字符串 <code>part</code> 都被删除：</p>
+
+<ul>
+	<li>找到 <code>s</code> 中 <strong>最左边</strong> 的子字符串 <code>part</code> ，并将它从 <code>s</code> 中删除。</li>
+</ul>
+
+<p>请你返回从 <code>s</code> 中删除所有 <code>part</code> 子字符串以后得到的剩余字符串。</p>
+
+<p>一个 <strong>子字符串</strong> 是一个字符串中连续的字符序列。</p>
+
+<p> </p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre><b>输入：</b>s = "daabcbaabcbc", part = "abc"
+<b>输出：</b>"dab"
+<b>解释：</b>以下操作按顺序执行：
+- s = "da<strong>abc</strong>baabcbc" ，删除下标从 2 开始的 "abc" ，得到 s = "dabaabcbc" 。
+- s = "daba<strong>abc</strong>bc" ，删除下标从 4 开始的 "abc" ，得到 s = "dababc" 。
+- s = "dab<strong>abc</strong>" ，删除下标从 3 开始的 "abc" ，得到 s = "dab" 。
+此时 s 中不再含有子字符串 "abc" 。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre><b>输入：</b>s = "axxxxyyyyb", part = "xy"
+<b>输出：</b>"ab"
+<b>解释：</b>以下操作按顺序执行：
+- s = "axxx<strong>xy</strong>yyyb" ，删除下标从 4 开始的 "xy" ，得到 s = "axxxyyyb" 。
+- s = "axx<strong>xy</strong>yyb" ，删除下标从 3 开始的 "xy" ，得到 s = "axxyyb" 。
+- s = "ax<strong>xy</strong>yb" ，删除下标从 2 开始的 "xy" ，得到 s = "axyb" 。
+- s = "a<strong>xy</strong>b" ，删除下标从 1 开始的 "xy" ，得到 s = "ab" 。
+此时 s 中不再含有子字符串 "xy" 。
+</pre>
+
+<p> </p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 1000</code></li>
+	<li><code>1 &lt;= part.length &lt;= 1000</code></li>
+	<li><code>s</code>​​​​​​ 和 <code>part</code> 只包小写英文字母。</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/1900-1999/1910.Remove All Occurrences of a Substring/README_EN.md

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+# [1910. Remove All Occurrences of a Substring](https://leetcode.com/problems/remove-all-occurrences-of-a-substring)
+
+[中文文档](/solution/1900-1999/1910.Remove%20All%20Occurrences%20of%20a%20Substring/README.md)
+
+## Description
+
+<p>Given two strings <code>s</code> and <code>part</code>, perform the following operation on <code>s</code> until <strong>all</strong> occurrences of the substring <code>part</code> are removed:</p>
+
+<ul>
+	<li>Find the <strong>leftmost</strong> occurrence of the substring <code>part</code> and <strong>remove</strong> it from <code>s</code>.</li>
+</ul>
+
+<p>Return <code>s</code><em> after removing all occurrences of </em><code>part</code>.</p>
+
+<p>A <strong>substring</strong> is a contiguous sequence of characters in a string.</p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;daabcbaabcbc&quot;, part = &quot;abc&quot;
+<strong>Output:</strong> &quot;dab&quot;
+<strong>Explanation</strong>: The following operations are done:
+- s = &quot;da<strong><u>abc</u></strong>baabcbc&quot;, remove &quot;abc&quot; starting at index 2, so s = &quot;dabaabcbc&quot;.
+- s = &quot;daba<strong><u>abc</u></strong>bc&quot;, remove &quot;abc&quot; starting at index 4, so s = &quot;dababc&quot;.
+- s = &quot;dab<strong><u>abc</u></strong>&quot;, remove &quot;abc&quot; starting at index 3, so s = &quot;dab&quot;.
+Now s has no occurrences of &quot;abc&quot;.
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;axxxxyyyyb&quot;, part = &quot;xy&quot;
+<strong>Output:</strong> &quot;ab&quot;
+<strong>Explanation</strong>: The following operations are done:
+- s = &quot;axxx<strong><u>xy</u></strong>yyyb&quot;, remove &quot;xy&quot; starting at index 4 so s = &quot;axxxyyyb&quot;.
+- s = &quot;axx<strong><u>xy</u></strong>yyb&quot;, remove &quot;xy&quot; starting at index 3 so s = &quot;axxyyb&quot;.
+- s = &quot;ax<strong><u>xy</u></strong>yb&quot;, remove &quot;xy&quot; starting at index 2 so s = &quot;axyb&quot;.
+- s = &quot;a<strong><u>xy</u></strong>b&quot;, remove &quot;xy&quot; starting at index 1 so s = &quot;ab&quot;.
+Now s has no occurrences of &quot;xy&quot;.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 1000</code></li>
+	<li><code>1 &lt;= part.length &lt;= 1000</code></li>
+	<li><code>s</code>​​​​​​ and <code>part</code> consists of lowercase English letters.</li>
+</ul>
+
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+
+```
+
+### **Java**
+
+```java
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->