Sum of Subarray Minimums

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
 
Solution:

class Solution {
    public int sumSubarrayMins(int[] A) {
        long sum = 0;
        int mod = (int) 1e9 + 7;
        Deque<Integer> stack = new ArrayDeque();
        for (int i = 0; i <= A.length; i ++) {
            int curr = i == A.length ? 0 : A[i];
            while (!stack.isEmpty() && curr < A[stack.peekFirst()]) {
                int index = stack.pop();
                // the distance between element A[index] and its previous smaller element
                int left = index - (stack.isEmpty() ? -1 : stack.peekFirst());
                // the distance between element A[index] and its next smaller element
                int right = i - index;
                sum = (sum + A[index] * left * right) % mod;
            }
            stack.push(i);
        }
        return (int) (sum % mod);
    }
}