<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();
        vector<int> next_smaller(n);
        long sum = 0;
        stack<pair<int, int>> st;
        st.push(make_pair(0, arr[0]));
        for(int i=1; i<arr.size(); i++){
            int top_elem = st.top().second;
            int idx = st.top().first;
            while((arr[i] < top_elem) ){
                next_smaller[idx] = i;
                st.pop();
                if(st.empty()) break;
                top_elem = st.top().second;
                idx = st.top().first;
            }
            st.push(make_pair(i, arr[i]));
        } 

        while(!(st.empty())){
            int idx = st.top().first;
            next_smaller[idx] = n;
            st.pop();
        }
        long M = 1000000000 + 7;
        for(int i=0; i<arr.size(); i++){
            // start with index i
            // if next_smaller[i] == j arr[i] is smallest for (j-i) subarrays starting from i.
            int begin = i;
            int end = next_smaller[begin];
            while(begin != n){
                sum =( sum + ((end - begin) * arr[begin]) % M) % M;
                begin = end;
                if(begin !=n)
                    end = next_smaller[begin];
            }
        } 
        return sum;
    }
};
44100