<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>& nums) {
        int n=nums.size();

        vector<int> dp(n); //dp[i] deontes sum of subarrays ending at i
        stack<int> s;
        int ans=0,mod=1e9+7;
        for(int i=0;i<n;i++){
            while(!s.empty() and nums[i]<nums[s.top()]){
                s.pop();
            }

            if(!s.empty()){ // means there is an element less than nums[i]
                int prev= s.top();
                dp[i]= dp[prev] + nums[i]*(i-prev);
            }
            else{
                dp[i]= nums[i]*(i+1);
            }

            s.push(i);

            ans= (ans+dp[i])%mod;
        }

        return ans;
    }
};
// https://leetcode.com/problems/sum-of-subarray-minimums/solutions/4595290/c-python-dp-monotonic-stack-23ms-beats-100/
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