diff --git a/Python/sum-of-subarray-minimums.py b/Python/sum-of-subarray-minimums.py
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+++ b/Python/sum-of-subarray-minimums.py
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+# Time:  O(n)
+# Space: O(n)
+
+# Given an array of integers A, find the sum of min(B),
+# where B ranges over every (contiguous) subarray of A.
+#
+# Since the answer may be large, return the answer modulo 10^9 + 7.
+#
+# Example 1:
+#
+# Input: [3,1,2,4]
+# Output: 17
+# Explanation: 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.
+#
+# Note:
+# - 1 <= A.length <= 30000
+# - 1 <= A[i] <= 30000
+
+import itertools
+
+
+# Ascending stack solution
+class Solution(object):
+    def sumSubarrayMins(self, A):
+        """
+        :type A: List[int]
+        :rtype: int
+        """
+        M = 10**9 + 7
+
+        left, s1 = [0]*len(A), []
+        for i in xrange(len(A)):
+            count = 1
+            while s1 and s1[-1][0] > A[i]:
+                count += s1.pop()[1]
+            left[i] = count
+            s1.append([A[i], count])
+
+        right, s2 = [0]*len(A), []
+        for i in reversed(xrange(len(A))):
+            count = 1
+            while s2 and s2[-1][0] >= A[i]:
+                count += s2.pop()[1]
+            right[i] = count
+            s2.append([A[i], count])
+
+        return sum(a*l*r for a, l, r in itertools.izip(A, left, right)) % M