907. Sum of Subarray Minimums # 题目 # 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
这两段区间包括2在内的所有subarray的最小值都是2,其分别可以组成的subarry的个数是 len([3])和len([4,5]),左右区间合并在一起可以组成的subarray的个数是len([3])*len([4,5]),再加上2本身自己组成一个独立的subarray [2],其总数就是 len([3]) + len([4,5]) + len([3])*len([4,5]) ...
we'll see a O(n) solution. For each array element A[i], if we can find out how many subarrays have it as the minimum in O(1) time, we'll get the total sum in O(n) time. To do this
Can you solve this real interview question? Sum of Subarray Minimums - 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. Ex
祖传的手艺不想丢了,所以按顺序写一个leetcode的题解。计划每日两题,争取不卡题吧 907.子数组的最小值之和 力扣leetcode-cn.com/problems/sum-of-subarray-minimums/ 对于这个题目,我们可以先转换一下思路,将计算各个子数组的最小值之和,转变为对于数组中的每个数来说,这个数是多少个子数组的最小值?
https://leetcode.com/contest/weekly-contest-102/problems/sum-of-subarray-minimums/ """ class Solution(object): def sumSubarrayMins(self, A): """ :type A: List[int] :rtype: int """ mod = 10**9 + 7 # count left = [] right = [] # (num, count) _left = [] _right = [...
907. Sum of Subarray Minimums 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… 阅读全文 赞同 添加评论 ...
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dp[i] represents the maximum sum of subarray which ends in nums[i], and dp[i] = Math.max(nums[i], dp[i - 1] + nums[i]). and since we have to include nums[i] due to it’s on the defination of dp[i], and when dp[i-1]<0 we can just choose not to add it. so the...