feat: add solutions to lc problems: No.2670~2673

* No.2670.Find the Distinct Difference Array
* No.2671.Frequency Tracker
* No.2672.Number of Adjacent Elements With the Same Color
* No.2673.Make Costs of Paths Equal in a Binary Tree

Author: yanglbme

solution/1100-1199/1114.Print in Order/README_EN.md

@@ -109,7 +109,7 @@ class Foo {
     private Semaphore c = new Semaphore(0);
 
     public Foo() {
-        
+
     }
 
     public void first(Runnable printFirst) throws InterruptedException {

solution/1100-1199/1116.Print Zero Even Odd/README_EN.md

@@ -69,7 +69,7 @@ class ZeroEvenOdd:
         self.z = Semaphore(1)
         self.e = Semaphore(0)
         self.o = Semaphore(0)
-        
+
 	# printNumber(x) outputs "x", where x is an integer.
     def zero(self, printNumber: 'Callable[[int], None]') -> None:
         for i in range(self.n):
@@ -79,13 +79,13 @@ class ZeroEvenOdd:
                 self.o.release()
             else:
                 self.e.release()
-        
+
     def even(self, printNumber: 'Callable[[int], None]') -> None:
         for i in range(2, self.n + 1, 2):
             self.e.acquire()
             printNumber(i)
             self.z.release()
-        
+
     def odd(self, printNumber: 'Callable[[int], None]') -> None:
         for i in range(1, self.n + 1, 2):
             self.o.acquire()
@@ -101,7 +101,7 @@ class ZeroEvenOdd {
     private Semaphore z = new Semaphore(1);
     private Semaphore e = new Semaphore(0);
     private Semaphore o = new Semaphore(0);
-    
+
     public ZeroEvenOdd(int n) {
         this.n = n;
     }

solution/1100-1199/1117.Building H2O/README.md

@@ -98,7 +98,7 @@ class H2O {
     private Semaphore o = new Semaphore(0);
 
     public H2O() {
-        
+
     }
 
     public void hydrogen(Runnable releaseHydrogen) throws InterruptedException {

solution/1100-1199/1117.Building H2O/README_EN.md

@@ -84,7 +84,7 @@ class H2O {
     private Semaphore o = new Semaphore(0);
 
     public H2O() {
-        
+
     }
 
     public void hydrogen(Runnable releaseHydrogen) throws InterruptedException {

solution/1200-1299/1216.Valid Palindrome III/README.md

@@ -6,7 +6,7 @@
 
 <!-- 这里写题目描述 -->
 
-<p>给出一个字符串&nbsp;<code>s</code>&nbsp;和一个整数&nbsp;<code>k</code>，若这个字符串是不是一个「k&nbsp;<strong>回文</strong>&nbsp;」，则返回 <code>true</code> 。</p>
+<p>给出一个字符串&nbsp;<code>s</code>&nbsp;和一个整数&nbsp;<code>k</code>，若这个字符串是一个「k&nbsp;<strong>回文</strong>&nbsp;」，则返回 <code>true</code> 。</p>
 
 <p>如果可以通过从字符串中删去最多 <code>k</code> 个字符将其转换为回文，那么这个字符串就是一个「<strong>k</strong>&nbsp;<strong>回文</strong>&nbsp;」。</p>
 

solution/2600-2699/2668.Find Latest Salaries/README.md

@@ -86,11 +86,11 @@ SELECT
 	firstname,
 	lastname,
 	max( salary ) AS salary,
-	department_id 
+	department_id
 FROM
-	Salary 
+	Salary
 GROUP BY
-	emp_id 
+	emp_id
 ORDER BY
 	emp_id;
 ```

solution/2600-2699/2668.Find Latest Salaries/README_EN.md

@@ -80,11 +80,11 @@ SELECT
 	firstname,
 	lastname,
 	max( salary ) AS salary,
-	department_id 
+	department_id
 FROM
-	Salary 
+	Salary
 GROUP BY
-	emp_id 
+	emp_id
 ORDER BY
 	emp_id;
 ```

solution/2600-2699/2669.Count Artist Occurrences On Spotify Ranking List/README.md

@@ -67,11 +67,11 @@ Each row contains an id, track_name, and artist.
 # Write your MySQL query statement below
 SELECT
 	artist,
-	count( 1 ) AS occurrences 
+	count( 1 ) AS occurrences
 FROM
-	Spotify 
+	Spotify
 GROUP BY
-	artist 
+	artist
 ORDER BY
 	occurrences DESC,
 	artist;

solution/2600-2699/2669.Count Artist Occurrences On Spotify Ranking List/README_EN.md

@@ -61,11 +61,11 @@ Each row contains an id, track_name, and artist.
 # Write your MySQL query statement below
 SELECT
 	artist,
-	count( 1 ) AS occurrences 
+	count( 1 ) AS occurrences
 FROM
-	Spotify 
+	Spotify
 GROUP BY
-	artist 
+	artist
 ORDER BY
 	occurrences DESC,
 	artist;

solution/2600-2699/2670.Find the Distinct Difference Array/README.md

@@ -0,0 +1,191 @@
+# [2670. 找出不同元素数目差数组](https://leetcode.cn/problems/find-the-distinct-difference-array)
+
+[English Version](/solution/2600-2699/2670.Find%20the%20Distinct%20Difference%20Array/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> 开始的数组 <code>nums</code> ，数组长度为 <code>n</code> 。</p>
+
+<p><code>nums</code> 的 <strong>不同元素数目差</strong> 数组可以用一个长度为 <code>n</code> 的数组 <code>diff</code> 表示，其中 <code>diff[i]</code> 等于前缀 <code>nums[0, ..., i]</code> 中不同元素的数目 <strong>减去</strong> 后缀 <code>nums[i + 1, ..., n - 1]</code> 中不同元素的数目。</p>
+
+<p>返回<em> </em><code>nums</code> 的 <strong>不同元素数目差</strong> 数组。</p>
+
+<p>注意 <code>nums[i, ..., j]</code> 表示 <code>nums</code> 的一个从下标 <code>i</code> 开始到下标 <code>j</code> 结束的子数组（包含下标 <code>i</code> 和 <code>j</code> 对应元素）。特别需要说明的是，如果 <code>i &gt; j</code> ，则 <code>nums[i, ..., j]</code> 表示一个空子数组。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>nums = [1,2,3,4,5]
+<strong>输出：</strong>[-3,-1,1,3,5]
+<strong>解释：
+</strong>对于 i = 0，前缀中有 1 个不同的元素，而在后缀中有 4 个不同的元素。因此，diff[0] = 1 - 4 = -3 。
+对于 i = 1，前缀中有 2 个不同的元素，而在后缀中有 3 个不同的元素。因此，diff[1] = 2 - 3 = -1 。
+对于 i = 2，前缀中有 3 个不同的元素，而在后缀中有 2 个不同的元素。因此，diff[2] = 3 - 2 = 1 。
+对于 i = 3，前缀中有 4 个不同的元素，而在后缀中有 1 个不同的元素。因此，diff[3] = 4 - 1 = 3 。
+对于 i = 4，前缀中有 5 个不同的元素，而在后缀中有 0 个不同的元素。因此，diff[4] = 5 - 0 = 5 。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>nums = [3,2,3,4,2]
+<strong>输出：</strong>[-2,-1,0,2,3]
+<strong>解释：</strong>
+对于 i = 0，前缀中有 1 个不同的元素，而在后缀中有 3 个不同的元素。因此，diff[0] = 1 - 3 = -2 。
+对于 i = 1，前缀中有 2 个不同的元素，而在后缀中有 3 个不同的元素。因此，diff[1] = 2 - 3 = -1 。
+对于 i = 2，前缀中有 2 个不同的元素，而在后缀中有 2 个不同的元素。因此，diff[2] = 2 - 2 = 0 。
+对于 i = 3，前缀中有 3 个不同的元素，而在后缀中有 1 个不同的元素。因此，diff[3] = 3 - 1 = 2 。
+对于 i = 4，前缀中有 3 个不同的元素，而在后缀中有 0 个不同的元素。因此，diff[4] = 3 - 0 = 3 。 
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n == nums.length&nbsp;&lt;= 50</code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 50</code></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 distinctDifferenceArray(self, nums: List[int]) -> List[int]:
+        n = len(nums)
+        ans = [0] * n
+        for i in range(n):
+            a = len(set(nums[: i + 1]))
+            b = len(set(nums[i + 1 :]))
+            ans[i] = a - b
+        return ans
+```
+
+```python
+class Solution:
+    def distinctDifferenceArray(self, nums: List[int]) -> List[int]:
+        n = len(nums)
+        suf = [0] * (n + 1)
+        s = set()
+        for i in range(n - 1, -1, -1):
+            s.add(nums[i])
+            suf[i] = len(s)
+
+        s.clear()
+        ans = [0] * n
+        for i, x in enumerate(nums):
+            s.add(x)
+            ans[i] = len(s) - suf[i + 1]
+        return ans
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+class Solution {
+    public int[] distinctDifferenceArray(int[] nums) {
+        int n = nums.length;
+        int[] suf = new int[n + 1];
+        Set<Integer> s = new HashSet<>();
+        for (int i = n - 1; i >= 0; --i) {
+            s.add(nums[i]);
+            suf[i] = s.size();
+        }
+        s.clear();
+        int[] ans = new int[n];
+        for (int i = 0; i < n; ++i) {
+            s.add(nums[i]);
+            ans[i] = s.size() - suf[i + 1];
+        }
+        return ans;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    vector<int> distinctDifferenceArray(vector<int>& nums) {
+        int n = nums.size();
+        vector<int> suf(n + 1);
+        unordered_set<int> s;
+        for (int i = n - 1; i >= 0; --i) {
+            s.insert(nums[i]);
+            suf[i] = s.size();
+        }
+        s.clear();
+        vector<int> ans(n);
+        for (int i = 0; i < n; ++i) {
+            s.insert(nums[i]);
+            ans[i] = s.size() - suf[i + 1];
+        }
+        return ans;
+    }
+};
+```
+
+### **Go**
+
+```go
+func distinctDifferenceArray(nums []int) []int {
+	n := len(nums)
+	suf := make([]int, n+1)
+	s := map[int]bool{}
+	for i := n - 1; i >= 0; i-- {
+		s[nums[i]] = true
+		suf[i] = len(s)
+	}
+	ans := make([]int, n)
+	s = map[int]bool{}
+	for i, x := range nums {
+		s[x] = true
+		ans[i] = len(s) - suf[i+1]
+	}
+	return ans
+}
+```
+
+### **TypeScript**
+
+```ts
+function distinctDifferenceArray(nums: number[]): number[] {
+    const n = nums.length;
+    const suf: number[] = new Array(n + 1).fill(0);
+    const s: Set<number> = new Set();
+    for (let i = n - 1; i >= 0; --i) {
+        s.add(nums[i]);
+        suf[i] = s.size;
+    }
+    s.clear();
+    const ans: number[] = new Array(n);
+    for (let i = 0; i < n; ++i) {
+        s.add(nums[i]);
+        ans[i] = s.size - suf[i + 1];
+    }
+    return ans;
+}
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->