Number of Subarrays with Bounded Maximum
<p>Given an integer array <code>nums</code> and two integers <code>left</code> and <code>right</code>, return <em>the number of contiguous non-empty <strong>subarrays</strong> such that the value of the maximum array element in that subarray is in the range </em><code>[left, right]</code>.</p>

<p>The test cases are generated so that the answer will fit in a <strong>32-bit</strong> integer.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,1,4,3], left = 2, right = 3
<strong>Output:</strong> 3
<strong>Explanation:</strong> There are three subarrays that meet the requirements: [2], [2, 1], [3].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,9,2,5,6], left = 2, right = 8
<strong>Output:</strong> 7
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>0 &lt;= left &lt;= right &lt;= 10<sup>9</sup></code></li>
</ul>
Array; Two Pointers
class Solution:
    def numSubarrayBoundedMax(self, nums: List[int], left: int, right: int) -> int:
        def f(x):
            cnt = t = 0
            for v in nums:
                t = 0 if v > x else t + 1
                cnt += t
            return cnt

        return f(right) - f(left - 1)
