<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 n = nums.size();
        vector<int> nextInvalid(n);
        int lastInvalid = n;
        
        for ( int i = n - 1; i >= 0; i-- ) {
            if ( nums[i] > right ) {
                lastInvalid = i;
            }
            nextInvalid[i] = lastInvalid;
        }

        int ans = 0, lastValid = n;

        for ( int i = n - 1; i >= 0; i-- ) {
            if ( nums[i] <= right && nums[i] >= left ) {
                ans += (nextInvalid[i] - i);
                lastValid = i;
            }
            if ( nums[i] < left ) {
                if ( lastValid != n && lastValid < nextInvalid[i] ) {
                    ans += (nextInvalid[lastValid] - lastValid);
                }
            }
            // cout << nextInvalid[i] << " ";
        }

        return ans;
    }
};
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