<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) {
        /*
       // brute force generating all sub array and finding the mini and adding to total it causes the TLE O(N^2)

        int mod=1000000007;
        int n=arr.size();
        int total=0;
        for(int i=0;i<n;i++)
        {
            int mini=arr[i];
            for(int j=i;j<n;j++)
            {
                mini=min(arr[j],mini);
                total=((total+mini)%mod)%mod;
            }
        }
        return total%mod;
        */

         int mod=1000000007;
        int n=arr.size();

        int pse[n];
        int nse[n];

        stack<int> st;
        for(int i=0;i<n;i++)
        {
            while(!st.empty()&&arr[i]<arr[st.top()])
            {
                st.pop();
            }

            if(st.empty())
              pse[i]=-1;
            else
             pse[i]=st.top();

            st.push(i);
        }

        stack<int> st1;
        for(int i=n-1;i>=0;i--)
        {
            while(!st1.empty()&&arr[i]<=arr[st1.top()])
            {
                st1.pop();
            }

            if(st1.empty())
              nse[i]=n;
            else
             nse[i]=st1.top();

            st1.push(i);
        }

         long long  total=0;

         for(int i=0;i<n;i++)
         {
            long long l=i-pse[i];
            long long  r=nse[i]-i;

        total = (total + (l * r % mod) * arr[i] % mod) % mod;
           // total=(total+(l*r%mod)*(arr[i]%mod))%mod;
         }

         return total%mod;
    }
};
43000