<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>

class Solution {
public:
    int numSubarrayBoundedMax(vector<int>& nums, int left, int right) {
        int start = -1, end = -1, count = 0;
        
        for (int i = 0; i < nums.size(); i++) {
            // If current number is greater than 'right', reset the valid subarray window
            if (nums[i] > right) {
                start = i;
            }
            // If current number is within the range, update 'end'
            if (nums[i] >= left) {
                end = i;
            }
            // Count all subarrays ending at 'i' and starting after 'start'
            count += end - start;
        }
        
        return count;
    }
};

55000