907. Sum of Subarray Minimums

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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. 1 <= A.length <= 30000
2. 1 <= A[i] <= 30000

思路：求每个数所有的subarray，分别求出能往左右2边能扩的距离，用单调栈来求

class Solution:
    def sumSubarrayMins(self, a):
        """
        :type A: List[int]
        :rtype: int
        """
        n=len(a)
        left,right=[0]*n,[0]*n
        
        st=[]
        for i,v in enumerate(a):
            while st and a[st[-1]]>v:
                idx=st.pop()
                right[idx]=i-idx
            st.append(i)
        while st:
            idx=st.pop()
            right[idx]=n-idx
        
        a=a[::-1]    
        st=[]
        for i,v in enumerate(a):
            while st and a[st[-1]]>=v:
                idx=st.pop()
                left[idx]=i-idx
            st.append(i)
        while st:
            idx=st.pop()
            left[idx]=n-idx
        left=left[::-1]
        
#        print(left,right)
        
        a=a[::-1]
        res=0
        mod=10**9+7
        for i in range(n):
#            print(left[i]*right[i], a[i])
            res+=left[i]*right[i]*a[i]
            res%=mod
        return res