<p>Given an array of integers arr, find the sum of <code>min(b)</code>, where <code>b</code> ranges over every (contiguous) subarray of <code>arr</code>. Since the answer may be large, return the answer <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

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

<pre>
<strong>Input:</strong> arr = [3,1,2,4]
<strong>Output:</strong> 17
<strong>Explanation:</strong> 
Subarrays are [3], [1], [2], [4], [3,1], [1,2], [2,4], [3,1,2], [1,2,4], [3,1,2,4]. 
Minimums are 3, 1, 2, 4, 1, 1, 2, 1, 1, 1.
Sum is 17.
</pre>

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

<pre>
<strong>Input:</strong> arr = [11,81,94,43,3]
<strong>Output:</strong> 444
</pre>

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

<ul>
	<li><code>1 &lt;= arr.length &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= arr[i] &lt;= 3 * 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    int sumSubarrayMins(vector<int>& arr) {
        long sum = 0;
        long modulo = 1e9 + 7;
        for (int idx = 0; idx < arr.size(); idx++) {
            int left = idx - 1, right = idx + 1;
            while(left >= 0 && arr[left] >= arr[idx])
                left--;
            while(right < arr.size() && arr[right] > arr[idx])
                right++;
            long num_arr = (right - idx) * (idx - left);
            sum = (sum + (num_arr * arr[idx]) % modulo) % modulo;
        }
        return (int)(sum % modulo);
    }
};

/* 
    3 1 2 4
    
    3 => 1 x 1 = 1
    XX1YY => 3 x 3 = 9 possible subarray with minimum 1
    3 3 => if duplicate number exists, use only one array (extend left pointer only)
    l

    min([3]) = 3
    min([3, 1]) = 1
    min([3, 1, 2]) = 1
    min([3, 1, 2, 4]) = 1
 */
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