The number of subarrays where max(subarray) <= target is easier to calculate. The answer is count(right) - count(left - 1). count(bound) counts subarrays where no element exceeds bound.

1. Define count_under(bound):
   • Use a sliding window or a simple counter to find contiguous segments where elements are <= bound.
   • For a segment of length L, there are L*(L+1)//2 subarrays.
2. Return count_under(right) - count_under(left - 1).

• Time Complexity: O(N).
• Space Complexity: O(1).

def num_subarray_bounded_max(nums, left, right):
    def count(bound):
        res = 0
        curr = 0
        for x in nums:
            if x <= bound:
                curr += 1
                res += curr
            else:
                curr = 0
        return res
        
    return count(right) - count(left - 1)