<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 {
    static int  solveSubtask( const int * begin , const int * end , int left )
    {
        int     total = 0;

        // Find first starter (value that is at least `left').
        auto  start = std::find_if( begin , end , [=](auto v){ return v >= left; } );
        if( start != end )
        {
            total += (end - start) * (start - begin + 1);   // count all subarrays containing  start.
            total += solveSubtask( start + 1 , end , left );
        }
        return total;
    }
public:
    int numSubarrayBoundedMax(vector<int>& nums, int left, int right)
    {
        int     total = 0;
        for( int b , s = 0 ; s < nums.size() ; s = b + 1 )
        {
            // Find next breaker (value greater than right).
            for( b = s ; b < nums.size() && nums[b] <= right ; ++b ){};

            if( b > s )
            {
                std::cout <<"# " <<s <<" .. " <<b <<endl;
                total += solveSubtask( nums.data() + s , nums.data() + b , left );
            }
        }
        return total;        
    }
};
58000