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feat: add solutions to lc problems: No.2465~2468
* No.2465.Number of Distinct Averages * No.2466.Count Ways To Build Good Strings * No.2467.Most Profitable Path in a Tree * No.2468.Split Message Based on Limit
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‎solution/2400-2499/2465.Number of Distinct Averages/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：排序 + 哈希表**
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我们先对数组进行排序，然后遍历数组，将 $nums[i]$ 与 $nums[n-i-1]$ 求和后放入哈希表中，最后返回哈希表的大小即可。
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时间复杂度 $O(n\times \log n)$，空间复杂度 $O(n)$。其中 $n$ 为数组的长度。
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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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class Solution:
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def distinctAverages(self, nums: List[int]) -> int:
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n = len(nums)
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nums.sort()
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return len(set(nums[i] + nums[n - i - 1] for i in range(n >> 1)))
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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 distinctAverages(int[] nums) {
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Arrays.sort(nums);
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int n = nums.length;
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Set<Integer> s = new HashSet<>();
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for (int i = 0; i < n >> 1; ++i) {
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s.add(nums[i] + nums[n - i - 1]);
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}
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return s.size();
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int distinctAverages(vector<int>& nums) {
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sort(nums.begin(), nums.end());
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int n = nums.size();
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unordered_set<int> s;
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for (int i = 0; i < n >> 1; ++i) s.insert(nums[i] + nums[n - i - 1]);
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return s.size();
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}
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};
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```
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### **Go**
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```go
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func distinctAverages(nums []int) int {
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sort.Ints(nums)
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n := len(nums)
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s := map[int]struct{}{}
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for i := 0; i < n>>1; i++ {
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s[nums[i]+nums[n-i-1]] = struct{}{}
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}
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return len(s)
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}
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```
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### **TypeScript**

‎solution/2400-2499/2465.Number of Distinct Averages/README_EN.md

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@@ -62,13 +62,56 @@ There is only one average to be calculated after removing 1 and 100, so we retur
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### **Python3**
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```python
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class Solution:
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def distinctAverages(self, nums: List[int]) -> int:
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n = len(nums)
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nums.sort()
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return len(set(nums[i] + nums[n - i - 1] for i in range(n >> 1)))
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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 distinctAverages(int[] nums) {
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Arrays.sort(nums);
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int n = nums.length;
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Set<Integer> s = new HashSet<>();
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for (int i = 0; i < n >> 1; ++i) {
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s.add(nums[i] + nums[n - i - 1]);
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}
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return s.size();
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int distinctAverages(vector<int>& nums) {
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sort(nums.begin(), nums.end());
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int n = nums.size();
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unordered_set<int> s;
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for (int i = 0; i < n >> 1; ++i) s.insert(nums[i] + nums[n - i - 1]);
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return s.size();
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}
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};
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```
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### **Go**
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```go
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func distinctAverages(nums []int) int {
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sort.Ints(nums)
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n := len(nums)
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s := map[int]struct{}{}
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for i := 0; i < n>>1; i++ {
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s[nums[i]+nums[n-i-1]] = struct{}{}
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}
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return len(s)
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}
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```
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### **TypeScript**

‎solution/2400-2499/2465.Number of Distinct Averages/Solution.cpp

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class Solution {
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public:
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int distinctAverages(vector<int>& nums) {
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sort(nums.begin(), nums.end());
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int n = nums.size();
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unordered_set<int> s;
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for (int i = 0; i < n >> 1; ++i) s.insert(nums[i] + nums[n - i - 1]);
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return s.size();
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}
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};

‎solution/2400-2499/2465.Number of Distinct Averages/Solution.go

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func distinctAverages(nums []int) int {
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sort.Ints(nums)
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n := len(nums)
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s := map[int]struct{}{}
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for i := 0; i < n>>1; i++ {
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s[nums[i]+nums[n-i-1]] = struct{}{}
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}
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return len(s)
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}

‎solution/2400-2499/2465.Number of Distinct Averages/Solution.java

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class Solution {
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public int distinctAverages(int[] nums) {
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Arrays.sort(nums);
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int n = nums.length;
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Set<Integer> s = new HashSet<>();
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for (int i = 0; i < n >> 1; ++i) {
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s.add(nums[i] + nums[n - i - 1]);
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}
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return s.size();
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}
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}

‎solution/2400-2499/2465.Number of Distinct Averages/Solution.py

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class Solution:
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def distinctAverages(self, nums: List[int]) -> int:
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n = len(nums)
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nums.sort()
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return len(set(nums[i] + nums[n - i - 1] for i in range(n >> 1)))

‎solution/2400-2499/2466.Count Ways To Build Good Strings/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：记忆化搜索**
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我们设计一个函数 $dfs(i)$ 表示从第 $i$ 位开始构造的好字符串的个数，答案即为 $dfs(0)$。
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函数 $dfs(i)$ 的计算过程如下：
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- 如果 $i \gt high$，返回 $0$；
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- 如果 $ low \leq i \leq high$，答案累加 $1$，然后 $i$ 之后既可以添加 `zero` 个 $0$，也可以添加 `one` 个 $1$，因此答案累加上 $dfs(i + zero) + dfs(i + one)$。
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过程中，我们需要对答案取模，并且可以使用记忆化搜索减少重复计算。
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时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n = high$。
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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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class Solution:
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def countGoodStrings(self, low: int, high: int, zero: int, one: int) -> int:
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@cache
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def dfs(i):
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if i > high:
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return 0
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ans = 0
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if low <= i <= high:
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ans += 1
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ans += dfs(i + zero) + dfs(i + one)
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return ans % mod
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mod = 10**9 + 7
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return dfs(0)
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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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private static final int MOD = (int) 1e9 + 7;
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private int[] f;
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private int lo;
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private int hi;
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private int zero;
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private int one;
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public int countGoodStrings(int low, int high, int zero, int one) {
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f = new int[high + 1];
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Arrays.fill(f, -1);
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lo = low;
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hi = high;
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this.zero = zero;
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this.one = one;
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return dfs(0);
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}
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private int dfs(int i) {
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if (i > hi) {
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return 0;
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}
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if (f[i] != -1) {
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return f[i];
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}
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long ans = 0;
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if (i >= lo && i <= hi) {
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++ans;
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}
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ans += dfs(i + zero) + dfs(i + one);
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ans %= MOD;
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f[i] = (int) ans;
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return f[i];
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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const int mod = 1e9 + 7;
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int countGoodStrings(int low, int high, int zero, int one) {
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vector<int> f(high + 1, -1);
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function<int(int)> dfs = [&](int i) -> int {
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if (i > high) return 0;
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if (f[i] != -1) return f[i];
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long ans = i >= low && i <= high;
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ans += dfs(i + zero) + dfs(i + one);
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ans %= mod;
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f[i] = ans;
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return ans;
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};
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return dfs(0);
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}
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};
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```
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### **Go**
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```go
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func countGoodStrings(low int, high int, zero int, one int) int {
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f := make([]int, high+1)
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for i := range f {
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f[i] = -1
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}
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const mod int = 1e9 + 7
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var dfs func(i int) int
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dfs = func(i int) int {
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if i > high {
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return 0
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}
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if f[i] != -1 {
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return f[i]
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}
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ans := 0
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if i >= low && i <= high {
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ans++
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}
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ans += dfs(i+zero) + dfs(i+one)
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ans %= mod
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f[i] = ans
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return ans
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}
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return dfs(0)
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}
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```
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### **TypeScript**

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