Link: https://leetcode.com/problems/count-complete-subarrays-in-an-array/description/
## 思路
sliding window
计算window里有多少个distinct number
如果数目等于```nums```里面的distinct number数 说明现在window里就是一个complete subarray 那么以左指针为起点 右指针和右指针后面的每一个index为终点的subarray都是complete subarray
因此```ans += len(nums)-r```
接着移动左指针

也可以用brute force 最坏情况下的时间复杂度是一样的
## Code

class Solution:
    def countCompleteSubarrays(self, nums: List[int]) -> int:
        distinctInWhole = len(set(nums))
        currDistinct = 0
        count = collections.defaultdict(int)
        l, r = 0, 0
        ans = 0
        while r<len(nums):
            if count[nums[r]] == 0:
                currDistinct += 1
            count[nums[r]] += 1
            while l<=r and currDistinct==distinctInWhole:
                ans += len(nums)-r
                count[nums[l]] -= 1
                if count[nums[l]] == 0:
                    currDistinct -= 1
                l += 1
            r += 1
        return ans