<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) {
        int n = arr.size();
        const int mod = 1e9+7;
        vector<int> left(n, -1);
        vector<int> right(n, n);
        vector<int> stack;

        // Find previous less element
        for (int i = 0; i < n; ++i) {
            while (!stack.empty() && arr[stack.back()] >= arr[i]) {
                stack.pop_back();
            }
            left[i] = stack.empty() ? -1 : stack.back();
            stack.push_back(i);
        }

        stack.clear();

        // Find next less element
        for (int i = n-1; i >= 0; i--) {
            while (!stack.empty() && arr[stack.back()] > arr[i]) {
                stack.pop_back();
            }
            right[i] = stack.empty() ? n : stack.back();
            stack.push_back(i);
        }

        //Calculate the result using the contribution of each element
        long long res = 0;
        for (int i = 0; i < n; ++i) {
            long long l = ( i - left[i] );
            long long r = ( right[i] - i );
            long long m = ( l%mod * r%mod ) % mod; 
            res = (res + ( arr[i] * m ) % mod ) % mod;
        }

        return (int) res;
    }
};
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