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feat: add solutions to lc problems: No.2072~2076
* No.2072.The Winner University * No.2073.Time Needed to Buy Tickets * No.2074.Reverse Nodes in Even Length Groups * No.2075.Decode the Slanted Ciphertext * No.2076.Process Restricted Friend Requests
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‎solution/2000-2099/2067.Number of Equal Count Substrings/README.md

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# [2067. Number of Equal Count Substrings](https://leetcode-cn.com/problems/number-of-equal-count-substrings)
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[English Version](/solution/2000-2099/2067.Number%20of%20Equal%20Count%20Substrings/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>You are given a <strong>0-indexed</strong> string <code>s</code> consisting of only lowercase English letters, and an integer <code>count</code>. A <strong>substring</strong> of <code>s</code> is said to be an <strong>equal count substring</strong> if, for each <strong>unique</strong> letter in the substring, it appears exactly <code>count</code> times in the substring.</p>
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<p>Return <em>the number of <strong>equal count substrings</strong> in </em><code>s</code>.</p>
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<p>A <strong>substring</strong> is a contiguous non-empty sequence of characters within 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;aaabcbbcc&quot;, count = 3
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<strong>Output:</strong> 3
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<strong>Explanation:</strong>
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The substring that starts at index 0 and ends at index 2 is &quot;aaa&quot;.
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The letter &#39;a&#39; in the substring appears exactly 3 times.
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The substring that starts at index 3 and ends at index 8 is &quot;bcbbcc&quot;.
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The letters &#39;b&#39; and &#39;c&#39; in the substring appear exactly 3 times.
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The substring that starts at index 0 and ends at index 8 is &quot;aaabcbbcc&quot;.
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The letters &#39;a&#39;, &#39;b&#39;, and &#39;c&#39; in the substring appear exactly 3 times.
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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;abcd&quot;, count = 2
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<strong>Output:</strong> 0
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<strong>Explanation:</strong>
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The number of times each letter appears in s is less than count.
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Therefore, no substrings in s are equal count substrings, so 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;a&quot;, count = 5
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<strong>Output:</strong> 0
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<strong>Explanation:</strong>
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The number of times each letter appears in s is less than count.
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Therefore, no substrings in s are equal count substrings, so return 0</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;= 3 * 10<sup>4</sup></code></li>
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<li><code>1 &lt;= count &lt;= 3 * 10<sup>4</sup></code></li>
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<li><code>s</code> consists only of lowercase English letters.</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 -->

‎solution/2000-2099/2067.Number of Equal Count Substrings/README_EN.md

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# [2067. Number of Equal Count Substrings](https://leetcode.com/problems/number-of-equal-count-substrings)
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[中文文档](/solution/2000-2099/2067.Number%20of%20Equal%20Count%20Substrings/README.md)
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## Description
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<p>You are given a <strong>0-indexed</strong> string <code>s</code> consisting of only lowercase English letters, and an integer <code>count</code>. A <strong>substring</strong> of <code>s</code> is said to be an <strong>equal count substring</strong> if, for each <strong>unique</strong> letter in the substring, it appears exactly <code>count</code> times in the substring.</p>
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<p>Return <em>the number of <strong>equal count substrings</strong> in </em><code>s</code>.</p>
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<p>A <strong>substring</strong> is a contiguous non-empty sequence of characters within 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;aaabcbbcc&quot;, count = 3
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<strong>Output:</strong> 3
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<strong>Explanation:</strong>
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The substring that starts at index 0 and ends at index 2 is &quot;aaa&quot;.
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The letter &#39;a&#39; in the substring appears exactly 3 times.
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The substring that starts at index 3 and ends at index 8 is &quot;bcbbcc&quot;.
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The letters &#39;b&#39; and &#39;c&#39; in the substring appear exactly 3 times.
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The substring that starts at index 0 and ends at index 8 is &quot;aaabcbbcc&quot;.
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The letters &#39;a&#39;, &#39;b&#39;, and &#39;c&#39; in the substring appear exactly 3 times.
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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;abcd&quot;, count = 2
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<strong>Output:</strong> 0
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<strong>Explanation:</strong>
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The number of times each letter appears in s is less than count.
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Therefore, no substrings in s are equal count substrings, so 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;a&quot;, count = 5
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<strong>Output:</strong> 0
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<strong>Explanation:</strong>
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The number of times each letter appears in s is less than count.
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Therefore, no substrings in s are equal count substrings, so return 0</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;= 3 * 10<sup>4</sup></code></li>
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<li><code>1 &lt;= count &lt;= 3 * 10<sup>4</sup></code></li>
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<li><code>s</code> consists only 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 -->

‎solution/2000-2099/2068.Check Whether Two Strings are Almost Equivalent/README.md

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# [2068. 检查两个字符串是否几乎相等](https://leetcode-cn.com/problems/check-whether-two-strings-are-almost-equivalent)
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[English Version](/solution/2000-2099/2068.Check%20Whether%20Two%20Strings%20are%20Almost%20Equivalent/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>如果两个字符串 <code>word1</code>&nbsp;和 <code>word2</code>&nbsp;中从 <code>'a'</code>&nbsp;到 <code>'z'</code>&nbsp;每一个字母出现频率之差都 <strong>不超过</strong>&nbsp;<code>3</code>&nbsp;，那么我们称这两个字符串&nbsp;<code>word1</code> 和&nbsp;<code>word2</code> <strong>几乎相等</strong>&nbsp;。</p>
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<p>给你两个长度都为&nbsp;<code>n</code>&nbsp;的字符串&nbsp;<code>word1</code> 和&nbsp;<code>word2</code>&nbsp;，如果&nbsp;<code>word1</code>&nbsp;和&nbsp;<code>word2</code>&nbsp;<strong>几乎相等</strong>&nbsp;，请你返回&nbsp;<code>true</code>&nbsp;，否则返回&nbsp;<code>false</code>&nbsp;。</p>
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<p>一个字母 <code>x</code>&nbsp;的出现 <strong>频率</strong>&nbsp;指的是它在字符串中出现的次数。</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>word1 = "aaaa", word2 = "bccb"
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<b>输出：</b>false
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<b>解释：</b>字符串 "aaaa" 中有 4 个 'a' ，但是 "bccb" 中有 0 个 'a' 。
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两者之差为 4 ，大于上限 3 。
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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>word1 = "abcdeef", word2 = "abaaacc"
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<b>输出：</b>true
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<b>解释：</b>word1 和 word2 中每个字母出现频率之差至多为 3 ：
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- 'a' 在 word1 中出现了 1 次，在 word2 中出现了 4 次，差为 3 。
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- 'b' 在 word1 中出现了 1 次，在 word2 中出现了 1 次，差为 0 。
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- 'c' 在 word1 中出现了 1 次，在 word2 中出现了 2 次，差为 1 。
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- 'd' 在 word1 中出现了 1 次，在 word2 中出现了 0 次，差为 1 。
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- 'e' 在 word1 中出现了 2 次，在 word2 中出现了 0 次，差为 2 。
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- 'f' 在 word1 中出现了 1 次，在 word2 中出现了 0 次，差为 1 。
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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>word1 = "cccddabba", word2 = "babababab"
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<b>输出：</b>true
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<b>解释：</b>word1 和 word2 中每个字母出现频率之差至多为 3 ：
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- 'a' 在 word1 中出现了 2 次，在 word2 中出现了 4 次，差为 2 。
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- 'b' 在 word1 中出现了 2 次，在 word2 中出现了 5 次，差为 3 。
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- 'c' 在 word1 中出现了 3 次，在 word2 中出现了 0 次，差为 3 。
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- 'd' 在 word1 中出现了 2 次，在 word2 中出现了 0 次，差为 2 。
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</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>n == word1.length == word2.length</code></li>
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<li><code>1 &lt;= n &lt;= 100</code></li>
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<li><code>word1</code> 和&nbsp;<code>word2</code>&nbsp;都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 -->

‎solution/2000-2099/2068.Check Whether Two Strings are Almost Equivalent/README_EN.md

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# [2068. Check Whether Two Strings are Almost Equivalent](https://leetcode.com/problems/check-whether-two-strings-are-almost-equivalent)
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[中文文档](/solution/2000-2099/2068.Check%20Whether%20Two%20Strings%20are%20Almost%20Equivalent/README.md)
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## Description
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<p>Two strings <code>word1</code> and <code>word2</code> are considered <strong>almost equivalent</strong> if the differences between the frequencies of each letter from <code>&#39;a&#39;</code> to <code>&#39;z&#39;</code> between <code>word1</code> and <code>word2</code> is <strong>at most</strong> <code>3</code>.</p>
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<p>Given two strings <code>word1</code> and <code>word2</code>, each of length <code>n</code>, return <code>true</code> <em>if </em><code>word1</code> <em>and</em> <code>word2</code> <em>are <strong>almost equivalent</strong>, or</em> <code>false</code> <em>otherwise</em>.</p>
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<p>The <strong>frequency</strong> of a letter <code>x</code> is the number of times it occurs in the 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> word1 = &quot;aaaa&quot;, word2 = &quot;bccb&quot;
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<strong>Output:</strong> false
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<strong>Explanation:</strong> There are 4 &#39;a&#39;s in &quot;aaaa&quot; but 0 &#39;a&#39;s in &quot;bccb&quot;.
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The difference is 4, which is more than the allowed 3.
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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> word1 = &quot;abcdeef&quot;, word2 = &quot;abaaacc&quot;
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<strong>Output:</strong> true
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<strong>Explanation:</strong> The differences between the frequencies of each letter in word1 and word2 are at most 3:
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- &#39;a&#39; appears 1 time in word1 and 4 times in word2. The difference is 3.
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- &#39;b&#39; appears 1 time in word1 and 1 time in word2. The difference is 0.
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- &#39;c&#39; appears 1 time in word1 and 2 times in word2. The difference is 1.
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- &#39;d&#39; appears 1 time in word1 and 0 times in word2. The difference is 1.
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- &#39;e&#39; appears 2 times in word1 and 0 times in word2. The difference is 2.
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- &#39;f&#39; appears 1 time in word1 and 0 times in word2. The difference is 1.
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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> word1 = &quot;cccddabba&quot;, word2 = &quot;babababab&quot;
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<strong>Output:</strong> true
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<strong>Explanation:</strong> The differences between the frequencies of each letter in word1 and word2 are at most 3:
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- &#39;a&#39; appears 2 times in word1 and 4 times in word2. The difference is 2.
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- &#39;b&#39; appears 2 times in word1 and 5 times in word2. The difference is 3.
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- &#39;c&#39; appears 3 times in word1 and 0 times in word2. The difference is 3.
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- &#39;d&#39; appears 2 times in word1 and 0 times in word2. The difference is 2.
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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>n == word1.length == word2.length</code></li>
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<li><code>1 &lt;= n &lt;= 100</code></li>
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<li><code>word1</code> and <code>word2</code> consist only 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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