<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>

from typing import List

class Solution:
    def sumSubarrayMins(self, arr: List[int]) -> int:
        MOD = 10**9 + 7  # Define modulo constant
        
        pse = []
        stack = []

        # Compute Previous Smaller Element (PSE)
        for i in range(len(arr)):
            while stack and stack[-1][0] >= arr[i]:
                stack.pop()

            if not stack:
                pse.append(-1)
            else:
                pse.append(stack[-1][1])
            
            stack.append((arr[i], i))

        nse = []
        stack = []

        # Compute Next Smaller Element (NSE)
        for i in range(len(arr)-1, -1, -1):
            while stack and stack[-1][0] > arr[i]:
                stack.pop()

            if not stack:
                nse.append(len(arr))
            else:
                nse.append(stack[-1][1])

            stack.append((arr[i], i))

        nse = nse[::-1]  # Reverse the list to correct order

        total = 0

        # Calculate the total sum of minimums in all subarrays
        for i in range(len(arr)):
            left_count = i - pse[i]       # Number of elements to the left
            right_count = nse[i] - i      # Number of elements to the right
            total += arr[i] * left_count * right_count
            total %= MOD                  # Apply modulo to avoid overflow

        return total

22000