<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) {
        
        //a left array which stores the number  of subarrays to left 
        //in which the current element is the min in the subarrays
        
        int n=arr.size();
        vector<int> left(n);
        
        //stack stores the indices
        stack<int> st;
        for(int i=0;i<n;i++)
        {
            while(!st.empty() && arr[st.top()]>arr[i])
                st.pop();
            
            //get the length
            if(st.empty())
                left[i]=i;
            else
                left[i]=(i-(st.top()+1));
            st.push(i);
        }
        
        //clear the stack
        while(!st.empty())
            st.pop();
        
        long res=0;
        long curr;
        long cnt;
        int mod=1000000007;
        //for the right sub array
        for(int i=n-1;i>=0;i--)
        {
            while(!st.empty() && arr[st.top()]>=arr[i])
                st.pop();
            if(st.empty())
                curr=n-i-1;
            else
                curr=st.top()-i-1;
            cnt=left[i]+curr+(left[i]*curr)+1;
            res+=(cnt*(long)arr[i]);
            res%=mod;
            st.push(i);
        }
        return (int)(res%mod);
            
    }
};
44200