feat: add solutions to lc problems: No.2586~2589

* No.2586.Count the Number of Vowel Strings in Range
* No.2587.Rearrange Array to Maximize Prefix Score
* No.2588.Count the Number of Beautiful Subarrays
* No.2589.Minimum Time to Complete All Tasks

Author: yanglbme

lcci/17.08.Circus Tower/README_EN.md

@@ -19,13 +19,11 @@
 	<li><code>height.length == weight.length &lt;= 10000</code></li>
 </ul>
 
-
 ## Solutions
 
-
 <!-- tabs:start -->
-### **Python3**
 
+### **Python3**
 
 ```python
 class BinaryIndexedTree:

lcof2/剑指 Offer II 020. 回文子字符串的个数/README.md

@@ -71,7 +71,7 @@ class Solution:
                 cnt += 1
                 i, j = i - 1, j + 1
             return cnt
-        
+
         n = len(s)
         return sum(f(i, i) + f(i, i + 1) for i in range(n))
 ```

solution/0100-0199/0154.Find Minimum in Rotated Sorted Array II/README.md

@@ -48,7 +48,7 @@
 
 <p>&nbsp;</p>
 
-<p><strong>进阶：</strong>这道题与 <a href="https://leetcode.cn/problems/find-minimum-in-rotated-sorted-array/description/">寻找旋转排序数组中的最小值</a> 类似，但 <code>nums</code> 可能包含重复元素。允许重复会影响算法的时间复杂度吗？会如何影响，为什么？</p>
+<p><strong>进阶：</strong>这道题与 <a href="https://leetcode-cn.com/problems/find-minimum-in-rotated-sorted-array/description/">寻找旋转排序数组中的最小值</a> 类似，但 <code>nums</code> 可能包含重复元素。允许重复会影响算法的时间复杂度吗？会如何影响，为什么？</p>
 
 ## 解法
 

solution/0100-0199/0168.Excel Sheet Column Title/README.md

@@ -17,7 +17,7 @@ C -> 3
 ...
 Z -> 26
 AA -> 27
-AB -> 28
+AB -> 28 
 ...
 </pre>
 

solution/0100-0199/0168.Excel Sheet Column Title/README_EN.md

@@ -15,7 +15,7 @@ C -> 3
 ...
 Z -> 26
 AA -> 27
-AB -> 28
+AB -> 28 
 ...
 </pre>
 

solution/0500-0599/0535.Encode and Decode TinyURL/README_EN.md

@@ -28,7 +28,7 @@
 <strong>Explanation:</strong>
 Solution obj = new Solution();
 string tiny = obj.encode(url); // returns the encoded tiny url.
-string ans = obj.decode(tiny); // returns the original url after deconding it.
+string ans = obj.decode(tiny); // returns the original url after decoding it.
 </pre>
 
 <p>&nbsp;</p>

solution/1500-1599/1516.Move Sub-Tree of N-Ary Tree/README_EN.md

@@ -4,7 +4,7 @@
 
 ## Description
 
-<p>Given the <code>root</code> of an <a href="https://leetcode.com/explore/learn/card/n-ary-tree/">N-ary tree</a> of unique values, and two nodes of the tree <code>p</code> and <code>q</code>.</p>
+<p>Given the <code>root</code> of an <span data-keyword="keyword-slug">n-ary-tree</span> of unique values, and two nodes of the tree <code>p</code> and <code>q</code>.</p>
 
 <p>You should move the subtree of the node <code>p</code> to become a direct child of node <code>q</code>. If <code>p</code> is already a direct child of <code>q</code>, do not change anything. Node <code>p</code> <strong>must be</strong> the last child in the children list of node <code>q</code>.</p>
 

solution/2500-2599/2577.Minimum Time to Visit a Cell In a Grid/README_EN.md

@@ -26,7 +26,7 @@
 - at t = 4, we move to the cell (1,1). It is possible because grid[1][1] <= 4.
 - at t = 5, we move to the cell (1,2). It is possible because grid[1][2] <= 5.
 - at t = 6, we move to the cell (1,3). It is possible because grid[1][3] <= 6.
-- at t = 7, we move to the cell (2,3). It is possible because grid[1][3] <= 7.
+- at t = 7, we move to the cell (2,3). It is possible because grid[2][3] <= 7.
 The final time is 7. It can be shown that it is the minimum time possible.
 </pre>
 

solution/2500-2599/2581.Count Number of Possible Root Nodes/README.md

@@ -68,6 +68,7 @@
 	<li><code>u<sub>j</sub> != v<sub>j</sub></code></li>
 	<li><code>edges</code>&nbsp;表示一棵有效的树。</li>
 	<li><code>guesses[j]</code>&nbsp;是树中的一条边。</li>
+	<li><code>guesses</code>&nbsp;是唯一的。</li>
 	<li><code>0 &lt;= k &lt;= guesses.length</code></li>
 </ul>
 

solution/2500-2599/2586.Count the Number of Vowel Strings in Range/README.md

@@ -0,0 +1,141 @@
+# [2586. 统计范围内的元音字符串数](https://leetcode.cn/problems/count-the-number-of-vowel-strings-in-range)
+
+[English Version](/solution/2500-2599/2586.Count%20the%20Number%20of%20Vowel%20Strings%20in%20Range/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> 和两个整数：<code>left</code> 和 <code>right</code> 。</p>
+
+<p>如果字符串以元音字母开头并以元音字母结尾，那么该字符串就是一个 <strong>元音字符串</strong> ，其中元音字母是 <code>'a'</code>、<code>'e'</code>、<code>'i'</code>、<code>'o'</code>、<code>'u'</code> 。</p>
+
+<p>返回<em> </em><code>words[i]</code> 是元音字符串的数目，其中<em> </em><code>i</code> 在闭区间 <code>[left, right]</code> 内。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>words = ["are","amy","u"], left = 0, right = 2
+<strong>输出：</strong>2
+<strong>解释：</strong>
+- "are" 是一个元音字符串，因为它以 'a' 开头并以 'e' 结尾。
+- "amy" 不是元音字符串，因为它没有以元音字母结尾。
+- "u" 是一个元音字符串，因为它以 'u' 开头并以 'u' 结尾。
+在上述范围中的元音字符串数目为 2 。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>words = ["hey","aeo","mu","ooo","artro"], left = 1, right = 4
+<strong>输出：</strong>3
+<strong>解释：</strong>
+- "aeo" 是一个元音字符串，因为它以 'a' 开头并以 'o' 结尾。
+- "mu" 不是元音字符串，因为它没有以元音字母开头。
+- "ooo" 是一个元音字符串，因为它以 'o' 开头并以 'o' 结尾。
+- "artro" 是一个元音字符串，因为它以 'a' 开头并以 'o' 结尾。
+在上述范围中的元音字符串数目为 3 。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= words.length &lt;= 1000</code></li>
+	<li><code>1 &lt;= words[i].length &lt;= 10</code></li>
+	<li><code>words[i]</code> 仅由小写英文字母组成</li>
+	<li><code>0 &lt;= left &lt;= right &lt; words.length</code></li>
+</ul>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+**方法一：模拟**
+
+我们只需要遍历区间 `[left, right]` 内的字符串，判断其是否以元音字母开头和结尾即可。若是，则答案加一。
+
+遍历结束后，返回答案即可。
+
+时间复杂度 $O(m)$，空间复杂度 $O(1)$。其中 $m = right - left + 1$。
+
+<!-- tabs:start -->
+
+### **Python3**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+class Solution:
+    def vowelStrings(self, words: List[str], left: int, right: int) -> int:
+        return sum(w[0] in 'aeiou' and w[-1] in 'aeiou' for w in words[left: right + 1])
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+class Solution {
+    public int vowelStrings(String[] words, int left, int right) {
+        int ans = 0;
+        for (int i = left; i <= right; ++i) {
+            var w = words[i];
+            if (check(w.charAt(0)) && check(w.charAt(w.length() - 1))) {
+                ++ans;
+            }
+        }
+        return ans;
+    }
+
+    private boolean check(char c) {
+        return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int vowelStrings(vector<string>& words, int left, int right) {
+        auto check = [](char c) -> bool {
+            return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
+        };
+        int ans = 0;
+        for (int i = left; i <= right; ++i) {
+            auto w = words[i];
+            ans += check(w[0]) && check(w[w.size() - 1]);
+        }
+        return ans;
+    }
+};
+```
+
+### **Go**
+
+```go
+func vowelStrings(words []string, left int, right int) (ans int) {
+	check := func(c byte) bool {
+		return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u'
+	}
+	for _, w := range words[left : right+1] {
+		if check(w[0]) && check(w[len(w)-1]) {
+			ans++
+		}
+	}
+	return
+}
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2500-2599/2586.Count the Number of Vowel Strings in Range/README_EN.md

@@ -0,0 +1,123 @@
+# [2586. Count the Number of Vowel Strings in Range](https://leetcode.com/problems/count-the-number-of-vowel-strings-in-range)
+
+[中文文档](/solution/2500-2599/2586.Count%20the%20Number%20of%20Vowel%20Strings%20in%20Range/README.md)
+
+## Description
+
+<p>You are given a <strong>0-indexed</strong> array of string <code>words</code> and two integers <code>left</code> and <code>right</code>.</p>
+
+<p>A string is called a <strong>vowel string</strong> if it starts with a vowel character and ends with a vowel character where vowel characters are <code>&#39;a&#39;</code>, <code>&#39;e&#39;</code>, <code>&#39;i&#39;</code>, <code>&#39;o&#39;</code>, and <code>&#39;u&#39;</code>.</p>
+
+<p>Return <em>the number of vowel strings </em><code>words[i]</code><em> where </em><code>i</code><em> belongs to the inclusive range </em><code>[left, right]</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> words = [&quot;are&quot;,&quot;amy&quot;,&quot;u&quot;], left = 0, right = 2
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> 
+- &quot;are&quot; is a vowel string because it starts with &#39;a&#39; and ends with &#39;e&#39;.
+- &quot;amy&quot; is not a vowel string because it does not end with a vowel.
+- &quot;u&quot; is a vowel string because it starts with &#39;u&#39; and ends with &#39;u&#39;.
+The number of vowel strings in the mentioned range is 2.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> words = [&quot;hey&quot;,&quot;aeo&quot;,&quot;mu&quot;,&quot;ooo&quot;,&quot;artro&quot;], left = 1, right = 4
+<strong>Output:</strong> 3
+<strong>Explanation:</strong> 
+- &quot;aeo&quot; is a vowel string because it starts with &#39;a&#39; and ends with &#39;o&#39;.
+- &quot;mu&quot; is not a vowel string because it does not start with a vowel.
+- &quot;ooo&quot; is a vowel string because it starts with &#39;o&#39; and ends with &#39;o&#39;.
+- &quot;artro&quot; is a vowel string because it starts with &#39;a&#39; and ends with &#39;o&#39;.
+The number of vowel strings in the mentioned range is 3.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= words.length &lt;= 1000</code></li>
+	<li><code>1 &lt;= words[i].length &lt;= 10</code></li>
+	<li><code>words[i]</code> consists of only lowercase English letters.</li>
+	<li><code>0 &lt;= left &lt;= right &lt; words.length</code></li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+class Solution:
+    def vowelStrings(self, words: List[str], left: int, right: int) -> int:
+        return sum(w[0] in 'aeiou' and w[-1] in 'aeiou' for w in words[left: right + 1])
+```
+
+### **Java**
+
+```java
+class Solution {
+    public int vowelStrings(String[] words, int left, int right) {
+        int ans = 0;
+        for (int i = left; i <= right; ++i) {
+            var w = words[i];
+            if (check(w.charAt(0)) && check(w.charAt(w.length() - 1))) {
+                ++ans;
+            }
+        }
+        return ans;
+    }
+
+    private boolean check(char c) {
+        return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int vowelStrings(vector<string>& words, int left, int right) {
+        auto check = [](char c) -> bool {
+            return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
+        };
+        int ans = 0;
+        for (int i = left; i <= right; ++i) {
+            auto w = words[i];
+            ans += check(w[0]) && check(w[w.size() - 1]);
+        }
+        return ans;
+    }
+};
+```
+
+### **Go**
+
+```go
+func vowelStrings(words []string, left int, right int) (ans int) {
+	check := func(c byte) bool {
+		return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u'
+	}
+	for _, w := range words[left : right+1] {
+		if check(w[0]) && check(w[len(w)-1]) {
+			ans++
+		}
+	}
+	return
+}
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2500-2599/2586.Count the Number of Vowel Strings in Range/Solution.cpp

@@ -0,0 +1,14 @@
+class Solution {
+public:
+    int vowelStrings(vector<string>& words, int left, int right) {
+        auto check = [](char c) -> bool {
+            return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';  
+        };
+        int ans = 0;
+        for (int i = left; i <= right; ++i) {
+            auto w = words[i];
+            ans += check(w[0]) && check(w[w.size() - 1]);
+        }
+        return ans;
+    }
+};

solution/2500-2599/2586.Count the Number of Vowel Strings in Range/Solution.go

@@ -0,0 +1,11 @@
+func vowelStrings(words []string, left int, right int) (ans int) {
+	check := func(c byte) bool {
+		return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u'
+	}
+	for _, w := range words[left : right+1] {
+		if check(w[0]) && check(w[len(w)-1]) {
+			ans++
+		}
+	}
+	return
+}

solution/2500-2599/2586.Count the Number of Vowel Strings in Range/Solution.java

@@ -0,0 +1,16 @@
+class Solution {
+    public int vowelStrings(String[] words, int left, int right) {
+        int ans = 0;
+        for (int i = left; i <= right; ++i) {
+            var w = words[i];
+            if (check(w.charAt(0)) && check(w.charAt(w.length() - 1))) {
+                ++ans;
+            }
+        }
+        return ans;
+    }
+    
+    private boolean check(char c) {
+        return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
+    }
+}