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Commit a75c20b

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feat: add new leetcode problems and solutions
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lines changed

40 files changed

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‎solution/1700-1799/1773.Count Items Matching a Rule/README.md‎

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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 countMatches(self, items: List[List[str]], ruleKey: str, ruleValue: str) -> int:
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count = 0
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m = {
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'type': 0,
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'color': 1,
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'name': 2
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}
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return sum([item[m[ruleKey]] == ruleValue for item in items])
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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 countMatches(List<List<String>> items, String ruleKey, String ruleValue) {
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int count = 0;
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for (List<String> item : items) {
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String t = item.get(0), c = item.get(1), n = item.get(2);
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if ("type".equals(ruleKey) && t.equals(ruleValue)) {
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++count;
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} else if ("color".equals(ruleKey) && c.equals(ruleValue)) {
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++count;
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} else if ("name".equals(ruleKey) && n.equals(ruleValue)) {
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++count;
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}
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}
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return count;
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}
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}
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```
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### **...**

‎solution/1700-1799/1773.Count Items Matching a Rule/README_EN.md‎

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### **Python3**
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```python
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class Solution:
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def countMatches(self, items: List[List[str]], ruleKey: str, ruleValue: str) -> int:
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count = 0
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m = {
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'type': 0,
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'color': 1,
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'name': 2
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}
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return sum([item[m[ruleKey]] == ruleValue for item in items])
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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 countMatches(List<List<String>> items, String ruleKey, String ruleValue) {
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int count = 0;
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for (List<String> item : items) {
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String t = item.get(0), c = item.get(1), n = item.get(2);
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if ("type".equals(ruleKey) && t.equals(ruleValue)) {
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++count;
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} else if ("color".equals(ruleKey) && c.equals(ruleValue)) {
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++count;
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} else if ("name".equals(ruleKey) && n.equals(ruleValue)) {
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++count;
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}
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}
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return count;
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}
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}
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```
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### **...**
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class Solution {
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public int countMatches(List<List<String>> items, String ruleKey, String ruleValue) {
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int count = 0;
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for (List<String> item : items) {
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String t = item.get(0), c = item.get(1), n = item.get(2);
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if ("type".equals(ruleKey) && t.equals(ruleValue)) {
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++count;
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} else if ("color".equals(ruleKey) && c.equals(ruleValue)) {
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++count;
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} else if ("name".equals(ruleKey) && n.equals(ruleValue)) {
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++count;
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}
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}
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return count;
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}
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}
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class Solution:
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def countMatches(self, items: List[List[str]], ruleKey: str, ruleValue: str) -> int:
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count = 0
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m = {
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'type': 0,
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'color': 1,
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'name': 2
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}
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return sum([item[m[ruleKey]] == ruleValue for item in items])
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# [1941. 检查是否所有字符出现次数相同](https://leetcode-cn.com/problems/check-if-all-characters-have-equal-number-of-occurrences)
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[English Version](/solution/1900-1999/1941.Check%20if%20All%20Characters%20Have%20Equal%20Number%20of%20Occurrences/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 <code>s</code> ,如果 <code>s</code> 是一个 <strong>好</strong> 字符串,请你返回 <code>true</code> ,否则请返回 <code>false</code> 。</p>
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<p>如果 <code>s</code> 中出现过的 <strong>所有</strong> 字符的出现次数 <strong>相同</strong> ,那么我们称字符串 <code>s</code> 是 <strong>好</strong> 字符串。</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 = "abacbc"
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<b>输出:</b>true
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<b>解释:</b>s 中出现过的字符为 'a','b' 和 'c' 。s 中所有字符均出现 2 次。
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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 = "aaabb"
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<b>输出:</b>false
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<b>解释:</b>s 中出现过的字符为 'a' 和 'b' 。
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'a' 出现了 3 次,'b' 出现了 2 次,两者出现次数不同。
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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>s</code> 只包含小写英文字母。</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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# [1941. Check if All Characters Have Equal Number of Occurrences](https://leetcode.com/problems/check-if-all-characters-have-equal-number-of-occurrences)
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[中文文档](/solution/1900-1999/1941.Check%20if%20All%20Characters%20Have%20Equal%20Number%20of%20Occurrences/README.md)
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## Description
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<p>Given a string <code>s</code>, return <code>true</code><em> if </em><code>s</code><em> is a <strong>good</strong> string, or </em><code>false</code><em> otherwise</em>.</p>
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<p>A string <code>s</code> is <strong>good</strong> if <strong>all</strong> the characters that appear in <code>s</code> have the <strong>same</strong> number of occurrences (i.e., the same frequency).</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;abacbc&quot;
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<strong>Output:</strong> true
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<strong>Explanation:</strong> The characters that appear in s are &#39;a&#39;, &#39;b&#39;, and &#39;c&#39;. All characters occur 2 times in s.
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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;aaabb&quot;
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<strong>Output:</strong> false
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<strong>Explanation:</strong> The characters that appear in s are &#39;a&#39; and &#39;b&#39;.
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&#39;a&#39; occurs 3 times while &#39;b&#39; occurs 2 times, which is not the same number of times.
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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>s</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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# [1942. 最小未被占据椅子的编号](https://leetcode-cn.com/problems/the-number-of-the-smallest-unoccupied-chair)
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[English Version](/solution/1900-1999/1942.The%20Number%20of%20the%20Smallest%20Unoccupied%20Chair/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>有 <code>n</code> 个朋友在举办一个派对,这些朋友从 <code>0</code> 到 <code>n - 1</code> 编号。派对里有 <strong>无数</strong> 张椅子,编号为 <code>0</code> 到 <code>infinity</code> 。当一个朋友到达派对时,他会占据 <strong>编号最小</strong> 且未被占据的椅子。</p>
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<ul>
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<li>比方说,当一个朋友到达时,如果椅子 <code>0</code> ,<code>1</code> 和 <code>5</code> 被占据了,那么他会占据 <code>2</code> 号椅子。</li>
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</ul>
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<p>当一个朋友离开派对时,他的椅子会立刻变成未占据状态。如果同一时刻有另一个朋友到达,可以立即占据这张椅子。</p>
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<p>给你一个下标从 <strong>0</strong> 开始的二维整数数组 <code>times</code> ,其中 <code>times[i] = [arrival<sub>i</sub>, leaving<sub>i</sub>]</code> 表示第 <code>i</code> 个朋友到达和离开的时刻,同时给你一个整数 <code>targetFriend</code> 。所有到达时间 <strong>互不相同</strong> 。</p>
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<p>请你返回编号为 <code>targetFriend</code> 的朋友占据的 <strong>椅子编号</strong> 。</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>times = [[1,4],[2,3],[4,6]], targetFriend = 1
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<b>输出:</b>1
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<b>解释:</b>
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- 朋友 0 时刻 1 到达,占据椅子 0 。
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- 朋友 1 时刻 2 到达,占据椅子 1 。
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- 朋友 1 时刻 3 离开,椅子 1 变成未占据。
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- 朋友 0 时刻 4 离开,椅子 0 变成未占据。
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- 朋友 2 时刻 4 到达,占据椅子 0 。
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朋友 1 占据椅子 1 ,所以返回 1 。
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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>times = [[3,10],[1,5],[2,6]], targetFriend = 0
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<b>输出:</b>2
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<b>解释:</b>
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- 朋友 1 时刻 1 到达,占据椅子 0 。
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- 朋友 2 时刻 2 到达,占据椅子 1 。
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- 朋友 0 时刻 3 到达,占据椅子 2 。
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- 朋友 1 时刻 5 离开,椅子 0 变成未占据。
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- 朋友 2 时刻 6 离开,椅子 1 变成未占据。
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- 朋友 0 时刻 10 离开,椅子 2 变成未占据。
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朋友 0 占据椅子 2 ,所以返回 2 。
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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>n == times.length</code></li>
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<li><code>2 &lt;= n &lt;= 10<sup>4</sup></code></li>
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<li><code>times[i].length == 2</code></li>
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<li><code>1 &lt;= arrival<sub>i</sub> &lt; leaving<sub>i</sub> &lt;= 10<sup>5</sup></code></li>
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<li><code>0 &lt;= targetFriend &lt;= n - 1</code></li>
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<li>每个 <code>arrival<sub>i</sub></code> 时刻 <strong>互不相同</strong> 。</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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