<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:
    '''
    Stack solution and dynamic programming: 

    dp[i] = sum of minima of all subarrays that end at index i precisely. 
    So that result will be sum(dp)

    How to calculate dp[i]. There are two cases:
    1. arr[i] is the smallest element so far. Then we have 
        dp[i] = (i+1)*arr[i]
    2. arr[i] is not the smallest element so far. But there's a previous smallest element at index prevsmaller. 
    Then for subarrays after prevsmaller the min is arr[i], whereas for subarrays that include prevsmaller, it's arr[prevsmaller].
    Recall that we are looking for the number of subarrays. So this is the same as number of subarrays at prevsmaller, whereas for the rest we have 
    i-prevsmaller subarrays. So we get
        dp[i] = dp[prevsmaller] + (i-prevsmaller)*arr[i].
    
    We need a way to keep track of the running minima. Note that we need here a monotonically increasing stack, because arr[prevsmaller]<arr[i]
    '''
    def sumSubarrayMins(self, arr: List[int]) -> int:
        
        MOD = 10**9+7
        dp = [0]*len(arr)
        stack = [] # index

        for i in range(len(arr)):

            while stack and arr[stack[-1]]>=arr[i]:
                stack.pop()
            
            if stack:
                # there is a prevsmaller element
                prevsmaller = stack[-1]
                dp[i] = (dp[prevsmaller] + (i-prevsmaller)*arr[i])%MOD
            else:
                dp[i] = (i+1)*arr[i]

            stack.append(i)

        return sum(dp)%MOD

21600