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feat: add solutions to lc problem: No.0992
No.0992.Subarrays with K Different Integers
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‎solution/0900-0999/0992.Subarrays with K Different Integers/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：双指针**
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我们遍历数组 $nums$，对于每个 $i$，我们需要找到最小的 $j_1$，使得 $nums[j_1], nums[j_1 + 1], \dots, nums[i]$ 中不同的整数个数小于等于 $k$，以及最小的 $j_2$，使得 $nums[j_2], nums[j_2 + 1], \dots, nums[i]$ 中不同的整数个数小于等于 $k-1$。那么 $j_2 - j_1$ 就是以 $i$ 结尾的满足条件的子数组的个数。
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在实现上，我们定义一个函数 $f(k)$，表示 $nums$ 中每个位置 $i$ 对应的最小的 $j$，使得 $nums[j], nums[j + 1], \dots, nums[i]$ 中不同的整数个数小于等于 $k$。该函数可以通过双指针实现，具体实现见代码。
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然后我们调用 $f(k)$ 和 $f(k-1)$，计算出每个位置对应的 $j_1$ 和 $j_2$，然后计算出以每个位置 $i$ 结尾的满足条件的子数组的个数，最后求和即可。
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时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $nums$ 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 subarraysWithKDistinct(self, nums: List[int], k: int) -> int:
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def f(k):
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pos = [0] * len(nums)
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cnt = Counter()
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j = 0
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for i, x in enumerate(nums):
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cnt[x] += 1
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while len(cnt) > k:
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cnt[nums[j]] -= 1
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if cnt[nums[j]] == 0:
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cnt.pop(nums[j])
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j += 1
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pos[i] = j
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return pos
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return sum(a - b for a, b in zip(f(k - 1), f(k)))
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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 subarraysWithKDistinct(int[] nums, int k) {
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int[] left = f(nums, k);
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int[] right = f(nums, k - 1);
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int ans = 0;
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for (int i = 0; i < nums.length; ++i) {
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ans += right[i] - left[i];
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}
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return ans;
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}
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private int[] f(int[] nums, int k) {
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int n = nums.length;
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int[] cnt = new int[n + 1];
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int[] pos = new int[n];
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int s = 0;
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for (int i = 0, j = 0; i < n; ++i) {
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if (++cnt[nums[i]] == 1) {
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++s;
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}
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for (; s > k; ++j) {
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if (--cnt[nums[j]] == 0) {
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--s;
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}
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}
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pos[i] = j;
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}
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return pos;
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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 subarraysWithKDistinct(vector<int>& nums, int k) {
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vector<int> left = f(nums, k);
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vector<int> right = f(nums, k - 1);
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int ans = 0;
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for (int i = 0; i < nums.size(); ++i) {
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ans += right[i] - left[i];
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}
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return ans;
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}
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vector<int> f(vector<int>& nums, int k) {
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int n = nums.size();
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vector<int> pos(n);
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int cnt[n + 1];
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memset(cnt, 0, sizeof(cnt));
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int s = 0;
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for (int i = 0, j = 0; i < n; ++i) {
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if (++cnt[nums[i]] == 1) {
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++s;
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}
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for (; s > k; ++j) {
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if (--cnt[nums[j]] == 0) {
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--s;
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}
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}
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pos[i] = j;
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}
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return pos;
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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 subarraysWithKDistinct(nums []int, k int) (ans int) {
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f := func(k int) []int {
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n := len(nums)
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pos := make([]int, n)
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cnt := make([]int, n+1)
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s, j := 0, 0
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for i, x := range nums {
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cnt[x]++
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if cnt[x] == 1 {
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s++
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}
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for ; s > k; j++ {
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cnt[nums[j]]--
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if cnt[nums[j]] == 0 {
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s--
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}
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}
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pos[i] = j
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}
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return pos
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}
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left, right := f(k), f(k-1)
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for i := range left {
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ans += right[i] - left[i]
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}
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return
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}
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```
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### **...**

‎solution/0900-0999/0992.Subarrays with K Different Integers/README_EN.md

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### **Python3**
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```python
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class Solution:
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def subarraysWithKDistinct(self, nums: List[int], k: int) -> int:
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def f(k):
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pos = [0] * len(nums)
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cnt = Counter()
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j = 0
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for i, x in enumerate(nums):
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cnt[x] += 1
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while len(cnt) > k:
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cnt[nums[j]] -= 1
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if cnt[nums[j]] == 0:
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cnt.pop(nums[j])
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j += 1
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pos[i] = j
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return pos
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return sum(a - b for a, b in zip(f(k - 1), f(k)))
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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 subarraysWithKDistinct(int[] nums, int k) {
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int[] left = f(nums, k);
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int[] right = f(nums, k - 1);
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int ans = 0;
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for (int i = 0; i < nums.length; ++i) {
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ans += right[i] - left[i];
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}
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return ans;
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}
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private int[] f(int[] nums, int k) {
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int n = nums.length;
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int[] cnt = new int[n + 1];
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int[] pos = new int[n];
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int s = 0;
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for (int i = 0, j = 0; i < n; ++i) {
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if (++cnt[nums[i]] == 1) {
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++s;
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}
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for (; s > k; ++j) {
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if (--cnt[nums[j]] == 0) {
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--s;
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}
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}
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pos[i] = j;
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}
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return pos;
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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 subarraysWithKDistinct(vector<int>& nums, int k) {
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vector<int> left = f(nums, k);
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vector<int> right = f(nums, k - 1);
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int ans = 0;
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for (int i = 0; i < nums.size(); ++i) {
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ans += right[i] - left[i];
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}
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return ans;
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}
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vector<int> f(vector<int>& nums, int k) {
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int n = nums.size();
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vector<int> pos(n);
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int cnt[n + 1];
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memset(cnt, 0, sizeof(cnt));
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int s = 0;
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for (int i = 0, j = 0; i < n; ++i) {
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if (++cnt[nums[i]] == 1) {
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++s;
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}
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for (; s > k; ++j) {
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if (--cnt[nums[j]] == 0) {
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--s;
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}
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}
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pos[i] = j;
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}
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return pos;
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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 subarraysWithKDistinct(nums []int, k int) (ans int) {
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f := func(k int) []int {
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n := len(nums)
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pos := make([]int, n)
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cnt := make([]int, n+1)
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s, j := 0, 0
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for i, x := range nums {
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cnt[x]++
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if cnt[x] == 1 {
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s++
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}
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for ; s > k; j++ {
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cnt[nums[j]]--
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if cnt[nums[j]] == 0 {
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s--
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}
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}
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pos[i] = j
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}
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return pos
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}
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left, right := f(k), f(k-1)
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for i := range left {
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ans += right[i] - left[i]
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}
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return
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}
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```
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### **...**

‎solution/0900-0999/0992.Subarrays with K Different Integers/Solution.cpp

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class Solution {
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public:
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int subarraysWithKDistinct(vector<int>& nums, int k) {
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vector<int> left = f(nums, k);
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vector<int> right = f(nums, k - 1);
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int ans = 0;
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for (int i = 0; i < nums.size(); ++i) {
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ans += right[i] - left[i];
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}
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return ans;
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}
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vector<int> f(vector<int>& nums, int k) {
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int n = nums.size();
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vector<int> pos(n);
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int cnt[n + 1];
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memset(cnt, 0, sizeof(cnt));
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int s = 0;
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for (int i = 0, j = 0; i < n; ++i) {
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if (++cnt[nums[i]] == 1) {
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++s;
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}
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for (; s > k; ++j) {
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if (--cnt[nums[j]] == 0) {
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--s;
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}
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}
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pos[i] = j;
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}
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return pos;
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}
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};

‎solution/0900-0999/0992.Subarrays with K Different Integers/Solution.go

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func subarraysWithKDistinct(nums []int, k int) (ans int) {
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f := func(k int) []int {
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n := len(nums)
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pos := make([]int, n)
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cnt := make([]int, n+1)
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s, j := 0, 0
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for i, x := range nums {
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cnt[x]++
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if cnt[x] == 1 {
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s++
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}
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for ; s > k; j++ {
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cnt[nums[j]]--
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if cnt[nums[j]] == 0 {
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s--
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}
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}
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pos[i] = j
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}
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return pos
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}
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left, right := f(k), f(k-1)
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for i := range left {
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ans += right[i] - left[i]
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}
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return
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}

‎solution/0900-0999/0992.Subarrays with K Different Integers/Solution.java

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class Solution {
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public int subarraysWithKDistinct(int[] nums, int k) {
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int[] left = f(nums, k);
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int[] right = f(nums, k - 1);
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int ans = 0;
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for (int i = 0; i < nums.length; ++i) {
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ans += right[i] - left[i];
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}
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return ans;
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}
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private int[] f(int[] nums, int k) {
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int n = nums.length;
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int[] cnt = new int[n + 1];
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int[] pos = new int[n];
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int s = 0;
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for (int i = 0, j = 0; i < n; ++i) {
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if (++cnt[nums[i]] == 1) {
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++s;
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}
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for (; s > k; ++j) {
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if (--cnt[nums[j]] == 0) {
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--s;
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}
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}
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pos[i] = j;
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}
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return pos;
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}
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}

‎solution/0900-0999/0992.Subarrays with K Different Integers/Solution.py

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class Solution:
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def subarraysWithKDistinct(self, nums: List[int], k: int) -> int:
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def f(k):
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pos = [0] * len(nums)
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cnt = Counter()
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j = 0
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for i, x in enumerate(nums):
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cnt[x] += 1
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while len(cnt) > k:
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cnt[nums[j]] -= 1
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if cnt[nums[j]] == 0:
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cnt.pop(nums[j])
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j += 1
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pos[i] = j
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return pos
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return sum(a - b for a, b in zip(f(k - 1), f(k)))

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

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**方法一：模拟**
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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。
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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。
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遍历结束后，返回答案即可。
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