Question

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

Solution

  • Method 1: Brutal force O(N^2) 
      class Solution {
      public int sumSubarrayMins(int[] A) {
          if(A == null || A.length == 0) return 0;
          long sum = 0;
          for(int i = 0; i < A.length; i++){
              int min = A[i];
              for(int j = i; j < A.length; j++){
                  min = Math.min(A[j], min);
                  sum += min;
              }
          }
          return (int)(sum % (Math.pow(10, 9) + 7));
      }
  }