feat: add solutions to lc problems: No.2540~2547

* No.2540.Minimum Common Value
* No.2541.Minimum Operations to Make Array Equal II
* No.2542.Maximum Subsequence Score
* No.2543.Check if Point Is Reachable
* No.2544.Alternating Digit Sum
* No.2545.Sort the Students by Their Kth Score
* No.2546.Apply Bitwise Operations to Make Strings Equal
* No.2547.Minimum Cost to Split an Array

Author: yanglbme

solution/0100-0199/0117.Populating Next Right Pointers in Each Node II/README_EN.md

@@ -109,7 +109,7 @@ class Solution:
             if prev:
                 prev.next = curr
             prev = curr
-        
+
         node = root
         while node:
             prev = next = None
@@ -133,7 +133,7 @@ class Node {
     public Node next;
 
     public Node() {}
-    
+
     public Node(int _val) {
         val = _val;
     }
@@ -185,7 +185,7 @@ class Node {
     public Node next;
 
     public Node() {}
-    
+
     public Node(int _val) {
         val = _val;
     }

solution/0200-0299/0222.Count Complete Tree Nodes/README.md

@@ -107,7 +107,7 @@ class Solution:
                 d += 1
                 root = root.left
             return d
-        
+
         if root is None:
             return 0
         left, right = depth(root.left), depth(root.right)

solution/0200-0299/0222.Count Complete Tree Nodes/README_EN.md

@@ -76,7 +76,7 @@ class Solution:
                 d += 1
                 root = root.left
             return d
-        
+
         if root is None:
             return 0
         left, right = depth(root.left), depth(root.right)

solution/0800-0899/0812.Largest Triangle Area/README.md

@@ -6,25 +6,33 @@
 
 <!-- 这里写题目描述 -->
 
-<p>给定包含多个点的集合，从其中取三个点组成三角形，返回能组成的最大三角形的面积。</p>
+<p>给你一个由 <strong>X-Y</strong> 平面上的点组成的数组 <code>points</code> ，其中 <code>points[i] = [x<sub>i</sub>, y<sub>i</sub>]</code> 。从其中取任意三个不同的点组成三角形，返回能组成的最大三角形的面积。与真实值误差在 <code>10<sup>-5</sup></code> 内的答案将会视为正确答案<strong>。</strong></p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0800-0899/0812.Largest%20Triangle%20Area/images/1027.png" style="height: 369px; width: 450px;" />
+<pre>
+<strong>输入：</strong>points = [[0,0],[0,1],[1,0],[0,2],[2,0]]
+<strong>输出：</strong>2.00000
+<strong>解释：</strong>输入中的 5 个点如上图所示，红色的三角形面积最大。
+</pre>
+
+<p><strong class="example">示例 2：</strong></p>
 
 <pre>
-<strong>示例:</strong>
-<strong>输入:</strong> points = [[0,0],[0,1],[1,0],[0,2],[2,0]]
-<strong>输出:</strong> 2
-<strong>解释:</strong> 
-这五个点如下图所示。组成的橙色三角形是最大的，面积为2。
+<strong>输入：</strong>points = [[1,0],[0,0],[0,1]]
+<strong>输出：</strong>0.50000
 </pre>
 
-<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0800-0899/0812.Largest%20Triangle%20Area/images/1027.png" style="height:328px; width:400px" /></p>
+<p>&nbsp;</p>
 
-<p><strong>注意: </strong></p>
+<p><strong>提示：</strong></p>
 
 <ul>
-	<li><code>3 &lt;= points.length &lt;= 50</code>.</li>
-	<li>不存在重复的点。</li>
-	<li>&nbsp;<code>-50 &lt;= points[i][j] &lt;= 50</code>.</li>
-	<li>结果误差值在&nbsp;<code>10^-6</code>&nbsp;以内都认为是正确答案。</li>
+	<li><code>3 &lt;= points.length &lt;= 50</code></li>
+	<li><code>-50 &lt;= x<sub>i</sub>, y<sub>i</sub> &lt;= 50</code></li>
+	<li>给出的所有点 <strong>互不相同</strong></li>
 </ul>
 
 ## 解法

solution/1800-1899/1833.Maximum Ice Cream Bars/README.md

@@ -10,34 +10,37 @@
 
 <p>商店中新到 <code>n</code> 支雪糕，用长度为 <code>n</code> 的数组 <code>costs</code> 表示雪糕的定价，其中 <code>costs[i]</code> 表示第 <code>i</code> 支雪糕的现金价格。Tony 一共有 <code>coins</code> 现金可以用于消费，他想要买尽可能多的雪糕。</p>
 
-<p>给你价格数组 <code>costs</code> 和现金量 <code>coins</code> ，请你计算并返回 Tony 用 <code>coins</code> 现金能够买到的雪糕的 <strong>最大数量</strong> 。</p>
-
 <p><strong>注意：</strong>Tony 可以按任意顺序购买雪糕。</p>
 
-<p> </p>
+<p>给你价格数组 <code>costs</code> 和现金量 <code>coins</code> ，请你计算并返回 Tony 用 <code>coins</code> 现金能够买到的雪糕的 <strong>最大数量</strong> 。</p>
+
+<p>&nbsp;</p>
 
 <p><strong>示例 1：</strong></p>
 
-<pre><strong>输入：</strong>costs = [1,3,2,4,1], coins = 7
+<pre>
+<strong>输入：</strong>costs = [1,3,2,4,1], coins = 7
 <strong>输出：</strong>4
 <strong>解释：</strong>Tony 可以买下标为 0、1、2、4 的雪糕，总价为 1 + 3 + 2 + 1 = 7
 </pre>
 
 <p><strong>示例 2：</strong></p>
 
-<pre><strong>输入：</strong>costs = [10,6,8,7,7,8], coins = 5
+<pre>
+<strong>输入：</strong>costs = [10,6,8,7,7,8], coins = 5
 <strong>输出：</strong>0
 <strong>解释：</strong>Tony 没有足够的钱买任何一支雪糕。
 </pre>
 
 <p><strong>示例 3：</strong></p>
 
-<pre><strong>输入：</strong>costs = [1,6,3,1,2,5], coins = 20
+<pre>
+<strong>输入：</strong>costs = [1,6,3,1,2,5], coins = 20
 <strong>输出：</strong>6
 <strong>解释：</strong>Tony 可以买下所有的雪糕，总价为 1 + 6 + 3 + 1 + 2 + 5 = 18 。
 </pre>
 
-<p> </p>
+<p>&nbsp;</p>
 
 <p><strong>提示：</strong></p>
 

solution/2500-2599/2540.Minimum Common Value/README.md

@@ -0,0 +1,134 @@
+# [2540. 最小公共值](https://leetcode.cn/problems/minimum-common-value)
+
+[English Version](/solution/2500-2599/2540.Minimum%20Common%20Value/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>给你两个整数数组&nbsp;<code>nums1</code> 和&nbsp;<code>nums2</code>&nbsp;，它们已经按非降序排序，请你返回两个数组的 <strong>最小公共整数</strong>&nbsp;。如果两个数组&nbsp;<code>nums1</code> 和&nbsp;<code>nums2</code>&nbsp;没有公共整数，请你返回&nbsp;<code>-1</code>&nbsp;。</p>
+
+<p>如果一个整数在两个数组中都 <strong>至少出现一次</strong>&nbsp;，那么这个整数是数组&nbsp;<code>nums1</code> 和&nbsp;<code>nums2</code>&nbsp;<strong>公共</strong>&nbsp;的。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre><b>输入：</b>nums1 = [1,2,3], nums2 = [2,4]
+<b>输出：</b>2
+<b>解释：</b>两个数组的最小公共元素是 2 ，所以我们返回 2 。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre><b>输入：</b>nums1 = [1,2,3,6], nums2 = [2,3,4,5]
+<b>输出：</b>2
+<b>解释：</b>两个数组中的公共元素是 2 和 3 ，2 是较小值，所以返回 2 。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums1.length, nums2.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= nums1[i], nums2[j] &lt;= 10<sup>9</sup></code></li>
+	<li><code>nums1</code> 和&nbsp;<code>nums2</code>&nbsp;都是 <strong>非降序</strong>&nbsp;的。</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
+class Solution:
+    def getCommon(self, nums1: List[int], nums2: List[int]) -> int:
+        i = j = 0
+        m, n = len(nums1), len(nums2)
+        while i < m and j < n:
+            if nums1[i] == nums2[j]:
+                return nums1[i]
+            if nums1[i] < nums2[j]:
+                i += 1
+            else:
+                j += 1
+        return -1
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+class Solution {
+    public int getCommon(int[] nums1, int[] nums2) {
+        int m = nums1.length, n = nums2.length;
+        for (int i = 0, j = 0; i < m && j < n;) {
+            if (nums1[i] == nums2[j]) {
+                return nums1[i];
+            }
+            if (nums1[i] < nums2[j]) {
+                ++i;
+            } else {
+                ++j;
+            }
+        }
+        return -1;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int getCommon(vector<int>& nums1, vector<int>& nums2) {
+        int m = nums1.size(), n = nums2.size();
+        for (int i = 0, j = 0; i < m && j < n;) {
+            if (nums1[i] == nums2[j]) {
+                return nums1[i];
+            }
+            if (nums1[i] < nums2[j]) {
+                ++i;
+            } else {
+                ++j;
+            }
+        }
+        return -1;
+    }
+};
+```
+
+### **Go**
+
+```go
+func getCommon(nums1 []int, nums2 []int) int {
+	m, n := len(nums1), len(nums2)
+	for i, j := 0, 0; i < m && j < n; {
+		if nums1[i] == nums2[j] {
+			return nums1[i]
+		}
+		if nums1[i] < nums2[j] {
+			i++
+		} else {
+			j++
+		}
+	}
+	return -1
+}
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2500-2599/2540.Minimum Common Value/README_EN.md

@@ -0,0 +1,126 @@
+# [2540. Minimum Common Value](https://leetcode.com/problems/minimum-common-value)
+
+[中文文档](/solution/2500-2599/2540.Minimum%20Common%20Value/README.md)
+
+## Description
+
+<p>Given two integer arrays <code>nums1</code> and <code>nums2</code>, sorted in non-decreasing order, return <em>the <strong>minimum integer common</strong> to both arrays</em>. If there is no common integer amongst <code>nums1</code> and <code>nums2</code>, return <code>-1</code>.</p>
+
+<p>Note that an integer is said to be <strong>common</strong> to <code>nums1</code> and <code>nums2</code> if both arrays have <strong>at least one</strong> occurrence of that integer.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums1 = [1,2,3], nums2 = [2,4]
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> The smallest element common to both arrays is 2, so we return 2.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums1 = [1,2,3,6], nums2 = [2,3,4,5]
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> There are two common elements in the array 2 and 3 out of which 2 is the smallest, so 2 is returned.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums1.length, nums2.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= nums1[i], nums2[j] &lt;= 10<sup>9</sup></code></li>
+	<li>Both <code>nums1</code> and <code>nums2</code> are sorted in <strong>non-decreasing</strong> order.</li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+class Solution:
+    def getCommon(self, nums1: List[int], nums2: List[int]) -> int:
+        i = j = 0
+        m, n = len(nums1), len(nums2)
+        while i < m and j < n:
+            if nums1[i] == nums2[j]:
+                return nums1[i]
+            if nums1[i] < nums2[j]:
+                i += 1
+            else:
+                j += 1
+        return -1
+```
+
+### **Java**
+
+```java
+class Solution {
+    public int getCommon(int[] nums1, int[] nums2) {
+        int m = nums1.length, n = nums2.length;
+        for (int i = 0, j = 0; i < m && j < n;) {
+            if (nums1[i] == nums2[j]) {
+                return nums1[i];
+            }
+            if (nums1[i] < nums2[j]) {
+                ++i;
+            } else {
+                ++j;
+            }
+        }
+        return -1;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int getCommon(vector<int>& nums1, vector<int>& nums2) {
+        int m = nums1.size(), n = nums2.size();
+        for (int i = 0, j = 0; i < m && j < n;) {
+            if (nums1[i] == nums2[j]) {
+                return nums1[i];
+            }
+            if (nums1[i] < nums2[j]) {
+                ++i;
+            } else {
+                ++j;
+            }
+        }
+        return -1;
+    }
+};
+```
+
+### **Go**
+
+```go
+func getCommon(nums1 []int, nums2 []int) int {
+	m, n := len(nums1), len(nums2)
+	for i, j := 0, 0; i < m && j < n; {
+		if nums1[i] == nums2[j] {
+			return nums1[i]
+		}
+		if nums1[i] < nums2[j] {
+			i++
+		} else {
+			j++
+		}
+	}
+	return -1
+}
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->