LeetCode 992. Subarrays with K Different Integers 滑动窗口 at most

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该博客介绍了如何利用滑动窗口和字典数据结构解决一个编程问题，即给定一个整数数组，找出包含恰好k个不同整数的子数组数量。通过维护一个字典来跟踪子数组内的元素及其出现次数，博主展示了如何计算满足条件的子数组个数。示例代码中展示了对于不同输入的解决方案，例如输入数组[1,2,1,2,3]和k=2时，返回7个符合条件的子数组。

Given an array nums of positive integers, call a (contiguous, not necessarily distinct) subarray of nums good if the number of different integers in that subarray is exactly k.

(For example, [1,2,3,1,2] has 3 different integers: 1, 2, and 3.)

Return the number of good subarrays of nums.

Example 1:

Input: nums = [1,2,1,2,3], k = 2
Output: 7
Explanation: Subarrays formed with exactly 2 different integers: [1,2], [2,1], [1,2], [2,3], [1,2,1], [2,1,2], [1,2,1,2].

Example 2:

Input: nums = [1,2,1,3,4], k = 3
Output: 3
Explanation: Subarrays formed with exactly 3 different integers: [1,2,1,3], [2,1,3], [1,3,4].

Note:

1 <= nums.length <= 20000
1 <= nums[i] <= nums.length
1 <= k <= nums.length

------------------

直接滑动不太容易，用at_most数组辅助一下：

from typing import List

from collections import defaultdict
class Solution:
    def subarraysWithKDistinct(self, nums: List[int], k: int) -> int:
        def at_most(nums, k):
            l, i, j, res = len(nums), 0, 0, 0
            dic = defaultdict(int)
            while (j < l):
                dic[nums[j]] += 1
                j += 1

                while (len(dic) > k and i < j):
                    dic[nums[i]] -= 1
                    if (dic[nums[i]] == 0):
                        dic.pop(nums[i])
                    i += 1
                #统计[i,j)范围内，也就是左闭区间，右开区间数的个数
                res = res + (j - i)
            return res
        return at_most(nums, k) - at_most(nums, k-1)

s = Solution()
print(s.subarraysWithKDistinct(nums = [1,2], k = 1))
print(s.subarraysWithKDistinct(nums = [1,2,1,2,3], k = 2))
print(s.subarraysWithKDistinct(nums = [1,2,1,3,4], k = 3))