feat: add new lc problems

Author: yanglbme

solution/1300-1399/1381.Design a Stack With Increment Operation/README.md

@@ -6,12 +6,12 @@
 
 <!-- 这里写题目描述 -->
 
-<p>请你设计一个支持下述操作的栈。</p>
+<p>请你设计一个支持对其元素进行增量操作的栈。</p>
 
 <p>实现自定义栈类 <code>CustomStack</code> ：</p>
 
 <ul>
-	<li><code>CustomStack(int maxSize)</code>：用 <code>maxSize</code> 初始化对象，<code>maxSize</code> 是栈中最多能容纳的元素数量，栈在增长到 <code>maxSize</code> 之后则不支持 <code>push</code> 操作。</li>
+	<li><code>CustomStack(int maxSize)</code>：用 <code>maxSize</code> 初始化对象，<code>maxSize</code> 是栈中最多能容纳的元素数量。</li>
 	<li><code>void push(int x)</code>：如果栈还未增长到 <code>maxSize</code> ，就将 <code>x</code> 添加到栈顶。</li>
 	<li><code>int pop()</code>：弹出栈顶元素，并返回栈顶的值，或栈为空时返回 <strong>-1</strong> 。</li>
 	<li><code>void inc(int k, int val)</code>：栈底的 <code>k</code> 个元素的值都增加 <code>val</code> 。如果栈中元素总数小于 <code>k</code> ，则栈中的所有元素都增加 <code>val</code> 。</li>
@@ -21,35 +21,34 @@
 
 <p><strong>示例：</strong></p>
 
-<pre><strong>输入：</strong>
-[&quot;CustomStack&quot;,&quot;push&quot;,&quot;push&quot;,&quot;pop&quot;,&quot;push&quot;,&quot;push&quot;,&quot;push&quot;,&quot;increment&quot;,&quot;increment&quot;,&quot;pop&quot;,&quot;pop&quot;,&quot;pop&quot;,&quot;pop&quot;]
+<pre>
+<strong>输入：</strong>
+["CustomStack","push","push","pop","push","push","push","increment","increment","pop","pop","pop","pop"]
 [[3],[1],[2],[],[2],[3],[4],[5,100],[2,100],[],[],[],[]]
 <strong>输出：</strong>
 [null,null,null,2,null,null,null,null,null,103,202,201,-1]
 <strong>解释：</strong>
-CustomStack customStack = new CustomStack(3); // 栈是空的 []
-customStack.push(1);                          // 栈变为 [1]
-customStack.push(2);                          // 栈变为 [1, 2]
-customStack.pop();                            // 返回 2 --&gt; 返回栈顶值 2，栈变为 [1]
-customStack.push(2);                          // 栈变为 [1, 2]
-customStack.push(3);                          // 栈变为 [1, 2, 3]
-customStack.push(4);                          // 栈仍然是 [1, 2, 3]，不能添加其他元素使栈大小变为 4
-customStack.increment(5, 100);                // 栈变为 [101, 102, 103]
-customStack.increment(2, 100);                // 栈变为 [201, 202, 103]
-customStack.pop();                            // 返回 103 --&gt; 返回栈顶值 103，栈变为 [201, 202]
-customStack.pop();                            // 返回 202 --&gt; 返回栈顶值 202，栈变为 [201]
-customStack.pop();                            // 返回 201 --&gt; 返回栈顶值 201，栈变为 []
-customStack.pop();                            // 返回 -1 --&gt; 栈为空，返回 -1
+CustomStack stk = new CustomStack(3); // 栈是空的 []
+stk.push(1);                          // 栈变为 [1]
+stk.push(2);                          // 栈变为 [1, 2]
+stk.pop();                            // 返回 2 --&gt; 返回栈顶值 2，栈变为 [1]
+stk.push(2);                          // 栈变为 [1, 2]
+stk.push(3);                          // 栈变为 [1, 2, 3]
+stk.push(4);                          // 栈仍然是 [1, 2, 3]，不能添加其他元素使栈大小变为 4
+stk.increment(5, 100);                // 栈变为 [101, 102, 103]
+stk.increment(2, 100);                // 栈变为 [201, 202, 103]
+stk.pop();                            // 返回 103 --&gt; 返回栈顶值 103，栈变为 [201, 202]
+stk.pop();                            // 返回 202 --&gt; 返回栈顶值 202，栈变为 [201]
+stk.pop();                            // 返回 201 --&gt; 返回栈顶值 201，栈变为 []
+stk.pop();                            // 返回 -1 --&gt; 栈为空，返回 -1
 </pre>
 
 <p>&nbsp;</p>
 
 <p><strong>提示：</strong></p>
 
 <ul>
-	<li><code>1 &lt;= maxSize &lt;= 1000</code></li>
-	<li><code>1 &lt;= x &lt;= 1000</code></li>
-	<li><code>1 &lt;= k &lt;= 1000</code></li>
+	<li><code>1 &lt;= maxSize, x, k &lt;= 1000</code></li>
 	<li><code>0 &lt;= val &lt;= 100</code></li>
 	<li>每种方法 <code>increment</code>，<code>push</code> 以及 <code>pop</code> 分别最多调用 <code>1000</code> 次</li>
 </ul>

solution/2200-2299/2204.Distance to a Cycle in Undirected Graph/README_EN.md

@@ -16,7 +16,7 @@
 <p><strong class="example">Example 1:</strong></p>
 <img src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2200-2299/2204.Distance%20to%20a%20Cycle%20in%20Undirected%20Graph/images/image-20220315154238-1.png" style="width: 350px; height: 237px;" />
 <pre>
-<strong>Input:</strong> n = 7, edges = [[1,2],[2,3],[3,4],[4,1],[0,1],[5,2],[6,5]]
+<strong>Input:</strong> n = 7, edges = [[1,2],[2,4],[4,3],[3,1],[0,1],[5,2],[6,5]]
 <strong>Output:</strong> [1,0,0,0,0,1,2]
 <strong>Explanation:</strong>
 The nodes 1, 2, 3, and 4 form the cycle.
@@ -32,7 +32,7 @@ The distance from 6 to 2 is 2.
 <p><strong class="example">Example 2:</strong></p>
 <img src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2200-2299/2204.Distance%20to%20a%20Cycle%20in%20Undirected%20Graph/images/image-20220315154634-1.png" style="width: 400px; height: 297px;" />
 <pre>
-<strong>Input:</strong> n = 9, edges = [[0,1],[1,2],[0,2],[2,6],[6,7],[6,8],[1,3],[3,4],[3,5]]
+<strong>Input:</strong> n = 9, edges = [[0,1],[1,2],[0,2],[2,6],[6,7],[6,8],[0,3],[3,4],[3,5]]
 <strong>Output:</strong> [0,0,0,1,2,2,1,2,2]
 <strong>Explanation:</strong>
 The nodes 0, 1, and 2 form the cycle.

solution/2500-2599/2506.Count Pairs Of Similar Strings/README.md

@@ -0,0 +1,97 @@
+# [2506. 统计相似字符串对的数目](https://leetcode.cn/problems/count-pairs-of-similar-strings)
+
+[English Version](/solution/2500-2599/2506.Count%20Pairs%20Of%20Similar%20Strings/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> 开始的字符串数组 <code>words</code> 。</p>
+
+<p>如果两个字符串由相同的字符组成，则认为这两个字符串 <strong>相似</strong> 。</p>
+
+<ul>
+	<li>例如，<code>"abca"</code> 和 <code>"cba"</code> 相似，因为它们都由字符 <code>'a'</code>、<code>'b'</code>、<code>'c'</code> 组成。</li>
+	<li>然而，<code>"abacba"</code> 和 <code>"bcfd"</code> 不相似，因为它们不是相同字符组成的。</li>
+</ul>
+
+<p>请你找出满足字符串&nbsp;<code>words[i]</code><em> </em>和<em> </em><code>words[j]</code> 相似的下标对<em> </em><code>(i, j)</code><em> </em>，并返回下标对的数目，其中 <code>0 &lt;= i &lt; j &lt;= word.length - 1</code> 。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre><strong>输入：</strong>words = ["aba","aabb","abcd","bac","aabc"]
+<strong>输出：</strong>2
+<strong>解释：</strong>共有 2 对满足条件：
+- i = 0 且 j = 1 ：words[0] 和 words[1] 只由字符 'a' 和 'b' 组成。 
+- i = 3 且 j = 4 ：words[3] 和 words[4] 只由字符 'a'、'b' 和 'c' 。 
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre><strong>输入：</strong>words = ["aabb","ab","ba"]
+<strong>输出：</strong>3
+<strong>解释：</strong>共有 3 对满足条件：
+- i = 0 且 j = 1 ：words[0] 和 words[1] 只由字符 'a' 和 'b' 组成。 
+- i = 0 且 j = 2 ：words[0] 和 words[2] 只由字符 'a' 和 'b' 组成。 
+- i = 1 且 j = 2 ：words[1] 和 words[2] 只由字符 'a' 和 'b' 组成。 
+</pre>
+
+<p><strong>示例 3：</strong></p>
+
+<pre><strong>输入：</strong>words = ["nba","cba","dba"]
+<strong>输出：</strong>0
+<strong>解释：</strong>不存在满足条件的下标对，返回 0 。</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= words.length &lt;= 100</code></li>
+	<li><code>1 &lt;= words[i].length &lt;= 100</code></li>
+	<li><code>words[i]</code> 仅由小写英文字母组成</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
+
+```
+
+### **C++**
+
+```cpp
+
+```
+
+### **Go**
+
+```go
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2500-2599/2506.Count Pairs Of Similar Strings/README_EN.md

@@ -0,0 +1,90 @@
+# [2506. Count Pairs Of Similar Strings](https://leetcode.com/problems/count-pairs-of-similar-strings)
+
+[中文文档](/solution/2500-2599/2506.Count%20Pairs%20Of%20Similar%20Strings/README.md)
+
+## Description
+
+<p>You are given a <strong>0-indexed</strong> string array <code>words</code>.</p>
+
+<p>Two strings are <strong>similar</strong> if they consist of the same characters.</p>
+
+<ul>
+	<li>For example, <code>&quot;abca&quot;</code> and <code>&quot;cba&quot;</code> are similar since both consist of characters <code>&#39;a&#39;</code>, <code>&#39;b&#39;</code>, and <code>&#39;c&#39;</code>.</li>
+	<li>However, <code>&quot;abacba&quot;</code> and <code>&quot;bcfd&quot;</code> are not similar since they do not consist of the same characters.</li>
+</ul>
+
+<p>Return <em>the number of pairs </em><code>(i, j)</code><em> such that </em><code>0 &lt;= i &lt; j &lt;= word.length - 1</code><em> and the two strings </em><code>words[i]</code><em> and </em><code>words[j]</code><em> are similar</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> words = [&quot;aba&quot;,&quot;aabb&quot;,&quot;abcd&quot;,&quot;bac&quot;,&quot;aabc&quot;]
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> There are 2 pairs that satisfy the conditions:
+- i = 0 and j = 1 : both words[0] and words[1] only consist of characters &#39;a&#39; and &#39;b&#39;. 
+- i = 3 and j = 4 : both words[3] and words[4] only consist of characters &#39;a&#39;, &#39;b&#39;, and &#39;c&#39;. 
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> words = [&quot;aabb&quot;,&quot;ab&quot;,&quot;ba&quot;]
+<strong>Output:</strong> 3
+<strong>Explanation:</strong> There are 3 pairs that satisfy the conditions:
+- i = 0 and j = 1 : both words[0] and words[1] only consist of characters &#39;a&#39; and &#39;b&#39;. 
+- i = 0 and j = 2 : both words[0] and words[2] only consist of characters &#39;a&#39; and &#39;b&#39;.
+- i = 1 and j = 2 : both words[1] and words[2] only consist of characters &#39;a&#39; and &#39;b&#39;.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> words = [&quot;nba&quot;,&quot;cba&quot;,&quot;dba&quot;]
+<strong>Output:</strong> 0
+<strong>Explanation:</strong> Since there does not exist any pair that satisfies the conditions, we return 0.</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= words.length &lt;= 100</code></li>
+	<li><code>1 &lt;= words[i].length &lt;= 100</code></li>
+	<li><code>words[i]</code> consist of only lowercase English letters.</li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+
+```
+
+### **Java**
+
+```java
+
+```
+
+### **C++**
+
+```cpp
+
+```
+
+### **Go**
+
+```go
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2500-2599/2507.Smallest Value After Replacing With Sum of Prime Factors/README.md

@@ -0,0 +1,87 @@
+# [2507. 使用质因数之和替换后可以取到的最小值](https://leetcode.cn/problems/smallest-value-after-replacing-with-sum-of-prime-factors)
+
+[English Version](/solution/2500-2599/2507.Smallest%20Value%20After%20Replacing%20With%20Sum%20of%20Prime%20Factors/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>给你一个正整数 <code>n</code> 。</p>
+
+<p>请你将 <code>n</code> 的值替换为 <code>n</code> 的 <strong>质因数</strong> 之和，重复这一过程。</p>
+
+<ul>
+	<li>注意，如果 <code>n</code> 能够被某个质因数多次整除，则在求和时，应当包含这个质因数同样次数。</li>
+</ul>
+
+<p>返回<em> </em><code>n</code><em> </em>可以取到的最小值。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre><strong>输入：</strong>n = 15
+<strong>输出：</strong>5
+<strong>解释：</strong>最开始，n = 15 。
+15 = 3 * 5 ，所以 n 替换为 3 + 5 = 8 。
+8 = 2 * 2 * 2 ，所以 n 替换为 2 + 2 + 2 = 6 。
+6 = 2 * 3 ，所以 n 替换为 2 + 3 = 5 。
+5 是 n 可以取到的最小值。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre><strong>输入：</strong>n = 3
+<strong>输出：</strong>3
+<strong>解释：</strong>最开始，n = 3 。
+3 是 n 可以取到的最小值。</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>2 &lt;= n &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
+
+```
+
+### **C++**
+
+```cpp
+
+```
+
+### **Go**
+
+```go
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->