Description
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
Solution
Python3
class Solution:
def sumSubarrayMins(self, arr: List[int]) -> int:
N = len(arr)
MOD = 10 ** 9 + 7
prevSmaller = [-1] * N
nextSmaller = [N] * N
res = 0
# construct nextSmaller array
stack = []
for i, x in enumerate(arr):
while stack and x < arr[stack[-1]]:
nextSmaller[stack.pop()] = i
stack.append(i)
# construct prevSmaller array
stack = []
for i in range(N - 1, -1, -1):
# less than or equal here to prevent double counting
while stack and arr[i] <= arr[stack[-1]]:
prevSmaller[stack.pop()] = i
stack.append(i)
for i, x in enumerate(arr):
# fix x as the smallest in the subarray
left = i - prevSmaller[i]
right = nextSmaller[i] - i
res += x * left * right
res %= MOD
return res