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docs: add new lc problems
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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)
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[English Version](/solution/1900-1999/1909.Remove%20One%20Element%20to%20Make%20the%20Array%20Strictly%20Increasing/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<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>
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<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>
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<p> </p>
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<p><strong>示例 1:</strong></p>
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<pre><b>输入:</b>nums = [1,2,<strong>10</strong>,5,7]
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<b>输出:</b>true
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<b>解释:</b>从 nums 中删除下标 2 处的 10 ,得到 [1,2,5,7] 。
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[1,2,5,7] 是严格递增的,所以返回 true 。
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</pre>
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<p><strong>示例 2:</strong></p>
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<pre><b>输入:</b>nums = [2,3,1,2]
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<b>输出:</b>false
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<b>解释:</b>
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[3,1,2] 是删除下标 0 处元素后得到的结果。
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[2,1,2] 是删除下标 1 处元素后得到的结果。
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[2,3,2] 是删除下标 2 处元素后得到的结果。
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[2,3,1] 是删除下标 3 处元素后得到的结果。
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没有任何结果数组是严格递增的,所以返回 false 。</pre>
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<p><strong>示例 3:</strong></p>
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<pre><b>输入:</b>nums = [1,1,1]
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<b>输出:</b>false
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<b>解释:</b>删除任意元素后的结果都是 [1,1] 。
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[1,1] 不是严格递增的,所以返回 false 。
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</pre>
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<p><strong>示例 4:</strong></p>
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<pre><b>输入:</b>nums = [1,2,3]
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<b>输出:</b>true
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<b>解释:</b>[1,2,3] 已经是严格递增的,所以返回 true 。
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</pre>
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<p> </p>
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<p><strong>提示:</strong></p>
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<ul>
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<li><code>2 &lt;= nums.length &lt;= 1000</code></li>
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<li><code>1 &lt;= nums[i] &lt;= 1000</code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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```
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### **...**
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```
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```
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<!-- tabs:end -->
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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)
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[中文文档](/solution/1900-1999/1909.Remove%20One%20Element%20to%20Make%20the%20Array%20Strictly%20Increasing/README.md)
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## Description
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<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>
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<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>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [1,2,<u>10</u>,5,7]
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<strong>Output:</strong> true
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<strong>Explanation:</strong> By removing 10 at index 2 from nums, it becomes [1,2,5,7].
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[1,2,5,7] is strictly increasing, so return true.
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [2,3,1,2]
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<strong>Output:</strong> false
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<strong>Explanation:</strong>
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[3,1,2] is the result of removing the element at index 0.
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[2,1,2] is the result of removing the element at index 1.
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[2,3,2] is the result of removing the element at index 2.
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[2,3,1] is the result of removing the element at index 3.
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No resulting array is strictly increasing, so return false.</pre>
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<p><strong>Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [1,1,1]
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<strong>Output:</strong> false
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<strong>Explanation:</strong> The result of removing any element is [1,1].
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[1,1] is not strictly increasing, so return false.
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</pre>
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<p><strong>Example 4:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [1,2,3]
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<strong>Output:</strong> true
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<strong>Explanation:</strong> [1,2,3] is already strictly increasing, so return true.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>2 &lt;= nums.length &lt;= 1000</code></li>
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<li><code>1 &lt;= nums[i] &lt;= 1000</code></li>
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</ul>
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## Solutions
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<!-- tabs:start -->
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### **Python3**
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```python
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```
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### **Java**
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```java
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```
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### **...**
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```
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```
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<!-- tabs:end -->
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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)
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[English Version](/solution/1900-1999/1910.Remove%20All%20Occurrences%20of%20a%20Substring/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<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>
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<ul>
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<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>
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</ul>
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<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>
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<p>一个 <strong>子字符串</strong> 是一个字符串中连续的字符序列。</p>
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<p> </p>
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<p><strong>示例 1:</strong></p>
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<pre><b>输入:</b>s = "daabcbaabcbc", part = "abc"
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<b>输出:</b>"dab"
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<b>解释:</b>以下操作按顺序执行:
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- s = "da<strong>abc</strong>baabcbc" ,删除下标从 2 开始的 "abc" ,得到 s = "dabaabcbc" 。
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- s = "daba<strong>abc</strong>bc" ,删除下标从 4 开始的 "abc" ,得到 s = "dababc" 。
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- s = "dab<strong>abc</strong>" ,删除下标从 3 开始的 "abc" ,得到 s = "dab" 。
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此时 s 中不再含有子字符串 "abc" 。
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</pre>
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<p><strong>示例 2:</strong></p>
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<pre><b>输入:</b>s = "axxxxyyyyb", part = "xy"
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<b>输出:</b>"ab"
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<b>解释:</b>以下操作按顺序执行:
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- s = "axxx<strong>xy</strong>yyyb" ,删除下标从 4 开始的 "xy" ,得到 s = "axxxyyyb" 。
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- s = "axx<strong>xy</strong>yyb" ,删除下标从 3 开始的 "xy" ,得到 s = "axxyyb" 。
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- s = "ax<strong>xy</strong>yb" ,删除下标从 2 开始的 "xy" ,得到 s = "axyb" 。
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- s = "a<strong>xy</strong>b" ,删除下标从 1 开始的 "xy" ,得到 s = "ab" 。
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此时 s 中不再含有子字符串 "xy" 。
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</pre>
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<p> </p>
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<p><strong>提示:</strong></p>
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<ul>
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<li><code>1 &lt;= s.length &lt;= 1000</code></li>
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<li><code>1 &lt;= part.length &lt;= 1000</code></li>
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<li><code>s</code>​​​​​​ 和 <code>part</code> 只包小写英文字母。</li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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```
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### **...**
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```
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```
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<!-- tabs:end -->
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# [1910. Remove All Occurrences of a Substring](https://leetcode.com/problems/remove-all-occurrences-of-a-substring)
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[中文文档](/solution/1900-1999/1910.Remove%20All%20Occurrences%20of%20a%20Substring/README.md)
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## Description
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<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>
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<ul>
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<li>Find the <strong>leftmost</strong> occurrence of the substring <code>part</code> and <strong>remove</strong> it from <code>s</code>.</li>
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</ul>
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<p>Return <code>s</code><em> after removing all occurrences of </em><code>part</code>.</p>
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<p>A <strong>substring</strong> is a contiguous sequence of characters in a string.</p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> s = &quot;daabcbaabcbc&quot;, part = &quot;abc&quot;
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<strong>Output:</strong> &quot;dab&quot;
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<strong>Explanation</strong>: The following operations are done:
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- s = &quot;da<strong><u>abc</u></strong>baabcbc&quot;, remove &quot;abc&quot; starting at index 2, so s = &quot;dabaabcbc&quot;.
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- s = &quot;daba<strong><u>abc</u></strong>bc&quot;, remove &quot;abc&quot; starting at index 4, so s = &quot;dababc&quot;.
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- s = &quot;dab<strong><u>abc</u></strong>&quot;, remove &quot;abc&quot; starting at index 3, so s = &quot;dab&quot;.
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Now s has no occurrences of &quot;abc&quot;.
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> s = &quot;axxxxyyyyb&quot;, part = &quot;xy&quot;
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<strong>Output:</strong> &quot;ab&quot;
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<strong>Explanation</strong>: The following operations are done:
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- s = &quot;axxx<strong><u>xy</u></strong>yyyb&quot;, remove &quot;xy&quot; starting at index 4 so s = &quot;axxxyyyb&quot;.
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- s = &quot;axx<strong><u>xy</u></strong>yyb&quot;, remove &quot;xy&quot; starting at index 3 so s = &quot;axxyyb&quot;.
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- s = &quot;ax<strong><u>xy</u></strong>yb&quot;, remove &quot;xy&quot; starting at index 2 so s = &quot;axyb&quot;.
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- s = &quot;a<strong><u>xy</u></strong>b&quot;, remove &quot;xy&quot; starting at index 1 so s = &quot;ab&quot;.
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Now s has no occurrences of &quot;xy&quot;.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= s.length &lt;= 1000</code></li>
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<li><code>1 &lt;= part.length &lt;= 1000</code></li>
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<li><code>s</code>​​​​​​ and <code>part</code> consists of lowercase English letters.</li>
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</ul>
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## Solutions
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<!-- tabs:start -->
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### **Python3**
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```python
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```
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### **Java**
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```java
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```
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### **...**
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```
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```
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<!-- tabs:end -->

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