At Most K - Helper Function

function atMost(nums, k):
    count = empty map
    left = 0
    result = 0

    for right from 0 to nums.length - 1:
        count[nums[right]] = count.get(nums[right], 0) + 1

        while count.size > k:
            count[nums[left]] = count[nums[left]] - 1
            if count[nums[left]] == 0:
                count.remove(nums[left])
            left = left + 1

        result = result + (right - left + 1)

    return result

Time: $O(n)$. Space: $O(k)$.