Sum of Subarray Minimums
<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>
Stack; Array; Dynamic Programming; Monotonic Stack
class Solution:
    def sumSubarrayMins(self, arr: List[int]) -> int:
        n = len(arr)
        left = [-1] * n
        right = [n] * n
        stk = []
        for i, v in enumerate(arr):
            while stk and arr[stk[-1]] >= v:
                stk.pop()
            if stk:
                left[i] = stk[-1]
            stk.append(i)

        stk = []
        for i in range(n - 1, -1, -1):
            while stk and arr[stk[-1]] > arr[i]:
                stk.pop()
            if stk:
                right[i] = stk[-1]
            stk.append(i)
        mod = 10**9 + 7
        return sum((i - left[i]) * (right[i] - i) * v for i, v in enumerate(arr)) % mod
