<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:
    vector<int> nextSmallerElement(vector<int>& arr){
        vector<int> res;
        stack<int> s;
        int n = arr.size();
        for(int i=n-1; i>=0; i--){
            while(!s.empty() && arr[s.top()] >= arr[i])
                s.pop();
            res.push_back(s.empty()? n : s.top());
            s.push(i);
        }
        reverse(res.begin(), res.end());
        return res;
    }
    vector<int> prevSmallerOrEqualElement(vector<int>& arr){
        vector<int> res;
        stack<int> s;
        int n = arr.size();
        for(int i=0; i<n; i++){
            while(!s.empty() && arr[s.top()] > arr[i])
                s.pop();
            res.push_back(s.empty()? -1: s.top());
            s.push(i);
        }
        return res;
    }
    int sumSubarrayMins(vector<int>& arr) {
        vector<int> nse = nextSmallerElement(arr);
        vector<int> pse = prevSmallerOrEqualElement(arr);
        for(int i=0; i<arr.size(); i++)
            cout<<nse[i]<<" ";
        cout<<endl;
        for(int i=0; i<arr.size(); i++)
            cout<<pse[i]<<" ";
        
        int sum = 0;
        int mod = 1000000007;
        for(int i=0; i<arr.size(); i++){
            //no. of subarrays for which arr[i] is the min
            int x = ((nse[i]-1) - i + 1) % mod;
            int y = (i - (pse[i] + 1) + 1) % mod;
            int subarrays = (((x*y*1LL)%mod)*arr[i])%mod;
            sum = (sum + subarrays%mod) % mod;
        }
        return sum;
    }
};
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