Problem URL: Count the number of good subarray

We will use a sliding window to count the number of good subarray.

class Solution:
    def countGood(self, nums: List[int], k: int) -> int:
        res = cur = i = 0
        count = Counter()
        for j in range(len(nums)):
            k -= count[nums[j]]
            count[nums[j]] += 1
            while k <= 0:
                count[nums[i]] -= 1
                k += count[nums[i]]
                i += 1
            res += i
        return res

Time complexity: O(n), n is the length of the array.
Space complexity: O(n)