2799. Count Complete Subarrays in an Array

link

For each occurrence, we just keep 1 time

When we are at i index and the counter == set number, it means we have all element for nums[left: i]

we can just add left+1 times

(nums[left-1: i], nums[left-2: i], nums[left-3: i], ... nums[0: i])

class Solution:
    def countCompleteSubarrays(self, nums: List[int]) -> int:

        n = len(nums)
        
        if n == 1:
            return 1
        
        S = set(nums)
        
        M = defaultdict(int)
        
        res = 0
        l = 0

        for i in range(n):
            M[nums[i]] += 1

            while l < i and M[nums[l]] > 1:
                M[nums[l]] -= 1
                l += 1
            if len(M.keys()) == len(S):
                res += l + 1

        return res