907. Sum of Subarray Minimums

https://leetcode.com/problems/sum-of-subarray-minimums/

Given an array of integers arr, find the sum of min(b), where b ranges over every (contiguous) subarray of arr. Since the answer may be large, return the answer modulo 109 + 7.

Example 1:

Input: arr = [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.

Example 2:

Input: arr = [11,81,94,43,3]
Output: 444

Constraints:

• 1 <= arr.length <= 3 * 104

• 1 <= arr[i] <= 3 * 104

问数组的所有子数组内最小值的和。sum(arr(i) * cnt(i)), cnt(i) 是以arr[i]为最小值的子数组个数。设[l, i, r]是以arr[i]为最小值的最长子数组，那cnt[i]即（左边选个长度 * 右边选个长度）的所有可能性: (i - l) * (r - i). 为了对每个arr[i]找到对应的l和r，用monotonic stack存当前最小值所在index，当arr[i]比s[-1]还小时，抛出s[-1]且res += (i - s[-1]) * (s[-1] * s[-2]), 在s.append(i)。

class Solution:
    def sumSubarrayMins(self, arr: List[int]) -> int:
        arr = [0] + arr + [0]
        s, res = [], 0
        for i, num in enumerate(arr):
            while s and num < arr[s[-1]]:
                j = s.pop()
                res = (res + (i - j) * (j - s[-1]) * arr[j]) % (10 ** 9 + 7)
            s.append(i)
        return res