LeetCode 907. Sum of Subarray Minimums
作者：互联网
原题链接在这里：https://leetcode.com/problems/sum-of-subarray-minimums/
题目：
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 * 1041 <= arr[i] <= 3 * 104
题解：
The minimum of subarray is straight forward. But how many subarrays containing this minimum value needs proof.
e.g. arr = [2, 9, 7, 8, 3, 4, 6, 1].
The element 3, there are 2 on its left, the distance is m = 4. There is 1 on its right, the distance is n = 3.
Overall, there are total m * n = 12 subarrays containing 3.
There are total m + n - 1 elements. The number of non-empty subarrays S3 is (1 + m + n - 1) * (m + n - 1) / 2.
On 3's left, [9, 8, 7]. The number of subarrays S1 = (1 + m - 1) * (m - 1) / 2.
One 3's right, [4, 6]. The number of subarrays S2 = (1 + n - 1) * (n - 1) / 2.
The number of subarrays containing 3 is S3 - S1 - S2 = m * n.
Have a monotonic increasing stack maintain the index, when encounter arr[i] is smaller stack top. Then we get the next smaller, like 1 in the example.
First pop we get the minimum elelment like 3 in the example.
If the stack is not empty, peek and we get the previous smaller like 2 in the example.
Then accumlate 3 * (left distance) * (right distance) to the sum.
Time Complexity: O(n). n = arr.length.
Space: O(n).
AC Java:
1 class Solution {
2 public int sumSubarrayMins(int[] arr) {
3 if(arr == null || arr.length == 0){
4 return 0;
5 }
6
7 long sum = 0;
8 Stack<Integer> stk = new Stack<>();
9 int n = arr.length;
10 for(int i = 0; i <= n; i++){
11 while(!stk.isEmpty() && arr[stk.peek()] > (i == n ? Integer.MIN_VALUE : arr[i])){
12 int minIndex = stk.pop();
13 int pre = stk.isEmpty() ? -1 : stk.peek();
14 sum += (long)arr[minIndex] * (i - minIndex) * (minIndex - pre);
15 }
16
17 stk.push(i);
18 }
19
20 return (int)(sum % (long)(1e9 + 7));
21 }
22 }
标签：minIndex,arr,Minimums,int,subarrays,Sum,stk,Subarray,sum 来源： https://www.cnblogs.com/Dylan-Java-NYC/p/16392011.html