Problem URL: Number of subarrays with bounded maximum

We will use two pointers for get the range. l is the left side of our sliding window, and r is the right side of our sliding window. We increment the size of our window (we increment r) if the encountered number is >= left. However, if the number is greater than right, we need to shrink the window to size 0, so we advance the left pointer to position i.

class Solution:
    def numSubarrayBoundedMax(self, nums: List[int], left: int, right: int) -> int:
        l, r, res = -1, -1, 0
        for i, num in enumerate(nums):
            if num >= left: r = i
            if num > right: l = i
            res += r - l
        return res

Time Complexity: O(n)
Space Complexity: O(1)