All subarrays : [1], [1, 2], [1, 2, 3], [2], [2, 3], [3] here first element 'arr[0]' appears 3 times second element 'arr[1]' appears 4 times third element 'arr[2]' appears 3 times Every element arr[i] appears in two types of subsets: i) In sybarrays beginning wi...
res = sum(A[i] * f(i)) 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...
No.907 Sum of Subarray Minimums 最小值子数组 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用单...
si是以Ai为尾的全部子列的sum(min)。如果A[n+1]<=A[n],s[n+1]=s[n]+A[n+1],否则假设A[n+1]左边第一个比它小的数是A[i],s[n+1] = s[i] + (n+1-i)*A[i]。 这样写可能更清晰,而且快: class Solution(object): def sumSubarrayMins(self, A): MOD = 10**9 + 7 N = len(...
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
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], ...
LeetCode Top 100 Liked Questions 560. Subarray Sum Equals K (Java版; Medium) 题目描述 AI检测代码解析 Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k.
Given an integer array nums and two integers firstLen and secondLen, return the maximum sum of elements in two non-overlapping subarrays with lengths firstLen and secondLen. The array with length firstLen could occur before or after the array with length secondLen, but they have to be non...
Can someone explain the optimal solution for finding the sum of XOR of all sub-arrays of the array. Example: Input : arr[] = {3, 8, 13} Output : 46 XOR of {3} = 3 XOR of {3, 8} = 11 XOR of {3, 8, 13} = 6 XOR of {8} = 8 XOR of {8, 13} = 5 XOR of {13...
if we encounter 1 atithithindex then for upcoming subarrays the behavior would become opposite (i.e. even 1s subarrays would act like odd 1s subarrays) wo we would dox=n−i−xx=n−i−xwheren−in−iis the number of subarrays starting withithithindex. ...