Medium
Sum of Subarray Minimums — Python
Full explanation · Time O(n) · Space O(n)
# Time: O(n)
# Space: O(n)
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