where f(i) is the number of subarrays, in which A[i] is the minimum. 难点就在于求f(i),为了求f(i)需要求left[i]和right[i]。 left[i]:A[i]左边严格大于A[i]的个数 right[i]:A[i]右边大于等于A[i]的个数 f(i) = (left[i] + 1) * (right[i] + 1),其实就是一个排列组合...
Given an array of integers arr, find the sum ofmin(b), wherebranges over every (contiguous) subarray ofarr. Since the answer may be large, return the answer modulo109 + 7. Example 1: Input:arr = [3,1,2,4]Output:17Explanation:Subarrays are [3], [1], [2], [4], [3,1], [...
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. Example 1: Input: arr = [3,1,2,4] Output: 17 Explanation: Subarrays are [3], [1], [2], ...
element of 3 element of 3 After finding both NLE and PLE of 3, we can determine the distance between 3 and 2(prevous less), and the distance between 3 and 1(next less). In this example, the distance is 4 and 3 respectively. How many subarray with 3 being its minimum value? The ...
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
907. Sum of Subarray Minimums 难度:m class Solution: def sumSubarrayMins(self, A: List[int]) -> int: res = 0 stack = [] for i, n in enumerate(A): while stack and A[stack[-1][0]] >= n: stack.pop() if stack: si = n*(i-stack[-1][0])+stack[-1][1] ...
No.907 Sum of Subarray Minimums 最小值子数组数组 做过较多类似题目的话很容易就想到枚举每个数字将其作为最小数时计算其所包含的子数组个数,要计算这样的子数组个数就会想到要计算以它为最小数的子数组的最大长度,自然就需要寻找它左右两边第一个比它小的数, 做过著名的Daily Temperatures那题的话很容易想到...
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
classSolution{public:intsumSubarrayMins(vector<int>&A){longlongsum=0;intn=A.size();vector<int>l(n,-1),r(n,n),st;for(inti=0;i<n;i++){while(!st.empty()&&A[st.back()]>A[i])st.pop_back();l[i]=st.empty()?-1:st.back();st.push_back(i);}st.clear();for(inti=n-...
No.907 Sum of Subarray Minimums 907. 子数组的最小值之和 - 力扣(LeetCode) (leetcode-cn.com) 思路参考:【LeetCode】907. Sum of Subarray Minimums_哔哩哔哩_bilibili 做过较多类似题目的话很容易就想到枚举每个数字将其作为最小数时计算其所包含的子数组个数,要计算这样的子数组个数就会想到要计算以它...