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feat: add solutions to lc problems: No.2315~2322
* No.2315.Count Asterisks * No.2316.Count Unreachable Pairs of Nodes in an Undirected Graph * No.2317.Maximum XOR After Operations * No.2318.Number of Distinct Roll Sequences * No.2319.Check if Matrix Is X-Matrix * No.2320.Count Number of Ways to Place Houses * No.2321.Maximum Score Of Spliced Array * No.2322.Minimum Score After Removals on a Tree
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‎solution/0700-0799/0710.Random Pick with Blacklist/Solution.py

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class Solution:
2-
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def __init__(self, n: int, blacklist: List[int]):
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self.k = n - len(blacklist)
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self.d = {}

‎solution/2300-2399/2315.Count Asterisks/README.md

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# [2315. 统计星号](https://leetcode.cn/problems/count-asterisks)
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[English Version](/solution/2300-2399/2315.Count%20Asterisks/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>给你一个字符串&nbsp;<code>s</code>&nbsp;，每&nbsp;<strong>两个</strong>&nbsp;连续竖线&nbsp;<code>'|'</code>&nbsp;为 <strong>一对</strong>&nbsp;。换言之，第一个和第二个&nbsp;<code>'|'</code>&nbsp;为一对，第三个和第四个&nbsp;<code>'|'</code>&nbsp;为一对，以此类推。</p>
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<p>请你返回 <strong>不在</strong> 竖线对之间，<code>s</code>&nbsp;中&nbsp;<code>'*'</code>&nbsp;的数目。</p>
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<p><strong>注意</strong>，每个竖线&nbsp;<code>'|'</code>&nbsp;都会 <strong>恰好</strong>&nbsp;属于一个对。</p>
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<p>&nbsp;</p>
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<p><strong>示例 1：</strong></p>
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<pre><b>输入：</b>s = "l|*e*et|c**o|*de|"
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<b>输出：</b>2
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<b>解释：</b>不在竖线对之间的字符加粗加斜体后，得到字符串："<strong><em>l</em></strong>|*e*et|<strong><em>c**o</em></strong>|*de|" 。
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第一和第二条竖线 '|' 之间的字符不计入答案。
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同时，第三条和第四条竖线 '|' 之间的字符也不计入答案。
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不在竖线对之间总共有 2 个星号，所以我们返回 2 。</pre>
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<p><strong>示例 2：</strong></p>
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<pre><b>输入：</b>s = "iamprogrammer"
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<b>输出：</b>0
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<b>解释：</b>在这个例子中，s 中没有星号。所以返回 0 。
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</pre>
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<p><strong>示例 3：</strong></p>
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<pre><b>输入：</b>s = "yo|uar|e**|b|e***au|tifu|l"
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<b>输出：</b>5
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<b>解释：</b>需要考虑的字符加粗加斜体后："<strong><em>yo</em></strong>|uar|<strong><em>e**</em></strong>|b|<strong><em>e***au</em></strong>|tifu|<strong><em>l</em></strong>" 。不在竖线对之间总共有 5 个星号。所以我们返回 5 。</pre>
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<p>&nbsp;</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>s</code>&nbsp;只包含小写英文字母，竖线&nbsp;<code>'|'</code>&nbsp;和星号&nbsp;<code>'*'</code>&nbsp;。</li>
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<li><code>s</code>&nbsp;包含 <strong>偶数</strong>&nbsp;个竖线&nbsp;<code>'|'</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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class Solution:
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def countAsterisks(self, s: str) -> int:
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ans = t = 0
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for c in s:
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if c == '|':
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t ^= 1
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elif c == '*':
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if t == 0:
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ans += 1
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return ans
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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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class Solution {
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public int countAsterisks(String s) {
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int ans = 0, t = 0;
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for (char c : s.toCharArray()) {
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if (c == '|') {
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t ^= 1;
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} else if (c == '*') {
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if (t == 0) {
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++ans;
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}
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int countAsterisks(string s) {
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int ans = 0, t = 0;
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for (char& c : s)
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{
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if (c == '|') t ^= 1;
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else if (c == '*') ans += t == 0;
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func countAsterisks(s string) int {
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ans, t := 0, 0
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for _, c := range s {
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if c == '|' {
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t ^= 1
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} else if c == '*' {
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if t == 0 {
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ans++
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}
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}
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}
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return ans
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}
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```
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### **TypeScript**
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```ts
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```
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### **...**
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```
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```
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<!-- tabs:end -->

‎solution/2300-2399/2315.Count Asterisks/README_EN.md

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# [2315. Count Asterisks](https://leetcode.com/problems/count-asterisks)
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[中文文档](/solution/2300-2399/2315.Count%20Asterisks/README.md)
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## Description
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<p>You are given a string <code>s</code>, where every <strong>two</strong> consecutive vertical bars <code>&#39;|&#39;</code> are grouped into a <strong>pair</strong>. In other words, the 1<sup>st</sup> and 2<sup>nd</sup> <code>&#39;|&#39;</code> make a pair, the 3<sup>rd</sup> and 4<sup>th</sup> <code>&#39;|&#39;</code> make a pair, and so forth.</p>
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<p>Return <em>the number of </em><code>&#39;*&#39;</code><em> in </em><code>s</code><em>, <strong>excluding</strong> the </em><code>&#39;*&#39;</code><em> between each pair of </em><code>&#39;|&#39;</code>.</p>
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<p><strong>Note</strong> that each <code>&#39;|&#39;</code> will belong to <strong>exactly</strong> one pair.</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;l|*e*et|c**o|*de|&quot;
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<strong>Output:</strong> 2
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<strong>Explanation:</strong> The considered characters are underlined: &quot;<u>l</u>|*e*et|<u>c**o</u>|*de|&quot;.
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The characters between the first and second &#39;|&#39; are excluded from the answer.
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Also, the characters between the third and fourth &#39;|&#39; are excluded from the answer.
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There are 2 asterisks considered. Therefore, we return 2.</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;iamprogrammer&quot;
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<strong>Output:</strong> 0
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<strong>Explanation:</strong> In this example, there are no asterisks in s. Therefore, we return 0.
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</pre>
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<p><strong>Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> s = &quot;yo|uar|e**|b|e***au|tifu|l&quot;
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<strong>Output:</strong> 5
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<strong>Explanation:</strong> The considered characters are underlined: &quot;<u>yo</u>|uar|<u>e**</u>|b|<u>e***au</u>|tifu|<u>l</u>&quot;. There are 5 asterisks considered. Therefore, we return 5.</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>s</code> consists of lowercase English letters, vertical bars <code>&#39;|&#39;</code>, and asterisks <code>&#39;*&#39;</code>.</li>
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<li><code>s</code> contains an <strong>even</strong> number of vertical bars <code>&#39;|&#39;</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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class Solution:
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def countAsterisks(self, s: str) -> int:
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ans = t = 0
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for c in s:
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if c == '|':
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t ^= 1
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elif c == '*':
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if t == 0:
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ans += 1
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return ans
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```
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### **Java**
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```java
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class Solution {
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public int countAsterisks(String s) {
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int ans = 0, t = 0;
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for (char c : s.toCharArray()) {
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if (c == '|') {
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t ^= 1;
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} else if (c == '*') {
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if (t == 0) {
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++ans;
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}
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int countAsterisks(string s) {
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int ans = 0, t = 0;
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for (char& c : s)
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{
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if (c == '|') t ^= 1;
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else if (c == '*') ans += t == 0;
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func countAsterisks(s string) int {
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ans, t := 0, 0
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for _, c := range s {
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if c == '|' {
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t ^= 1
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} else if c == '*' {
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if t == 0 {
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ans++
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}
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}
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}
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return ans
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}
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```
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### **TypeScript**
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```ts
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```
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### **...**
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```
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```
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<!-- tabs:end -->

‎solution/2300-2399/2315.Count Asterisks/Solution.cpp

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class Solution {
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public:
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int countAsterisks(string s) {
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int ans = 0, t = 0;
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for (char& c : s)
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{
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if (c == '|') t ^= 1;
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else if (c == '*') ans += t == 0;
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}
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return ans;
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}
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};

‎solution/2300-2399/2315.Count Asterisks/Solution.go

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func countAsterisks(s string) int {
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ans, t := 0, 0
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for _, c := range s {
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if c == '|' {
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t ^= 1
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} else if c == '*' {
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if t == 0 {
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ans++
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}
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}
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}
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return ans
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}

‎solution/2300-2399/2315.Count Asterisks/Solution.java

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class Solution {
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public int countAsterisks(String s) {
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int ans = 0, t = 0;
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for (char c : s.toCharArray()) {
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if (c == '|') {
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t ^= 1;
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} else if (c == '*') {
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if (t == 0) {
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++ans;
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}
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}
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}
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return ans;
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}
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}

‎solution/2300-2399/2315.Count Asterisks/Solution.py

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class Solution:
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def countAsterisks(self, s: str) -> int:
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ans = t = 0
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for c in s:
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if c == '|':
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t ^= 1
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elif c == '*':
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if t == 0:
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ans += 1
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return ans

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