907.Sum of Subarray Minimums
===

###### tags: `Medium`,`Array`,`DP`,`Stack`

[907. Sum of Subarray Minimums](https://leetcode.com/problems/sum-of-subarray-minimums/)

### 題目描述

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** $10^9 + 7$ .

### 範例

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.

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

**Constraints**:

* `1 <= arr.length` <= $3 * 10^4$
* `1 <= arr[i]` <= $3 * 10^4$

### 解答
#### Python

class Solution:
    def sumSubarrayMins(self, arr: List[int]) -> int:
        arr = [0] + arr
        stack = [0]
        dp = [0] * len(arr)

        for i, n in enumerate(arr):
            while arr[stack[-1]] > n:
                stack.pop()
            j = stack[-1]
            dp[i] = dp[j] + (i-j) * n
            stack.append(i)
        
        return sum(dp) % 1000000007

> [name=Yen-Chi Chen][time=Fri, Nov 25, 2022 10:40 PM]

class Solution:
    def sumSubarrayMins(self, arr: List[int]) -> int:
        res = 0
        stack = []
        arr = [0] + arr + [0]

        for idx, num in enumerate(arr):
            while stack and arr[stack[-1]] > num:
                cur = stack.pop()
                res += arr[cur] * (idx-cur) * (cur-stack[-1])
            stack.append(idx)
        
        return res% (10**9+7)

>[name=Dan][time=Sat, Jan 20, 2024 08:03 PM]

#### C++

class Solution {
public:
    int sumSubarrayMins(vector<int>& arr) {
        stack<int> s;
        vector<int> dp(arr.size());
        for (int i = 0; i < arr.size(); i++) {
            int n = arr[i];
            while(!s.empty() && arr[s.top()] >= n) {
                s.pop();
            }
            if (s.empty()) {
                dp[i] = (i + 1) * n;
            } else {
                int j = s.top();
                dp[i] = dp[j] + (i - j) * n;
            }
            s.push(i);
        }
        int result = 0;
        for (auto t : dp) {
            result = (result + t) % 1000000007;
        }
        return result;
    }
};

Time: $O(n)$
Extra Space: $O(n)$
> [name=Yen-Chi Chen][time=Fri, Nov 25, 2022 10:40 PM]

### Reference

[回到題目列表](https://hackmd.io/@Marsgoat/leetcode_every_day)