From b475dbf3ed59c94ebf85b1eea821b17f35307e07 Mon Sep 17 00:00:00 2001
From: gameloader <ggwqqo@163.com>
Date: Sat, 30 Mar 2024 11:27:20 +0800
Subject: [PATCH] leetcode update

---
 content/posts/leetcode.md | 90 +++++++++++++++++++++++++++++++++++++++
 1 file changed, 90 insertions(+)

diff --git a/content/posts/leetcode.md b/content/posts/leetcode.md
index 00d760b..9276122 100644
--- a/content/posts/leetcode.md
+++ b/content/posts/leetcode.md
@@ -2342,3 +2342,93 @@ func countSubarrays(nums []int, k int) int64 {
 	return count
 }
 ```
+
+## day32 2024-03-30
+
+### 992. Subarrays with K Different Integers
+
+Given an integer array nums and an integer k, return the number of good subarrays of nums.
+
+A good array is an array where the number of different integers in that array is exactly k.
+
+For example, [1,2,3,1,2] has 3 different integers: 1, 2, and 3.
+A subarray is a contiguous part of an array.
+![0330GvIAMJljJeoJ](https://testingcf.jsdelivr.net/gh/game-loader/picbase@master/uPic/0330GvIAMJljJeoJ.png)
+
+### 题解
+
+本题仍然使用滑动窗口, 这几天连续做滑动窗口的题, 总结了滑动窗口的"三要素": 1. 什么时候扩大窗口 2. 双么时候缩小窗口 3. 缩小和扩大窗口时执行哪些操作
+对于本题, 题目中要求计数正好包含k个不同数的子数组的个数, 求精确包含k个这种问题往往比较困难, 可以转化为求解包含≤k个和≤k-1个不同数的子数组个数的差. 这种思路求解起来非常方便. 对于求解包含≤k个不同数的子数组的个数, 当数组中包含不同数个数不足k时, 扩大窗口同时将计数增加当前窗口的长度, 若为k+1, 则缩小窗口至正好完全去除了一个数(若某个数只出现了1次, 去掉后窗口内不同数个数就是k, 若不止一次, 则去掉了不同数个数也没有变化, 故要继续向下遍历). 最后求≤k和≤k-1情况的计数值做差即可.
+
+### 代码
+
+```go
+func subarraysWithKDistinct(nums []int, k int) int {
+
+	return countk(nums, k) - countk(nums, k-1)
+}
+
+func countk(nums []int, k int) int {
+	count := map[int]int{}
+	different := 0
+	left := 0
+	result := 0
+	for index, value := range nums {
+		_, exist := count[value]
+		if !exist {
+			different++
+			count[value] = 1
+		} else {
+			count[value]++
+		}
+		if different <= k {
+			result += index - left + 1
+		}
+		if different == k+1 {
+			for count[nums[left]] > 1 {
+				count[nums[left]]--
+				left++
+			}
+			delete(count, nums[left])
+			left++
+			different--
+			result += index - left + 1
+		}
+	}
+    return result
+}
+```
+
+### 总结
+
+解题时没有注意题目限制, 后来查看最快解法发现忽略了题目中的数的范围, 题目中的数组中的数的大小不超过数组的长度, 数的范围已知, 因此可以使用数组代替哈希表来计数这样可以大大加快解题速度.
+
+```go
+func subarraysWithKDistinct(nums []int, k int) (ans int) {
+	f := func(k int) []int {
+		n := len(nums)
+		pos := make([]int, n)
+		cnt := make([]int, n+1)
+		s, j := 0, 0
+		for i, x := range nums {
+			cnt[x]++
+			if cnt[x] == 1 {
+				s++
+			}
+			for ; s > k; j++ {
+				cnt[nums[j]]--
+				if cnt[nums[j]] == 0 {
+					s--
+				}
+			}
+			pos[i] = j
+		}
+		return pos
+	}
+	left, right := f(k), f(k-1)
+	for i := range left {
+		ans += right[i] - left[i]
+	}
+	return
+}
+```