arr[i]is part of sub-array [1,2,3] but not part of [1] or [1, 2] (both starting from 1 which is part of preceded elements). Now the question is of how many subarrays of any preceded element,arr[i]is part of. Answer is(n-i)as subarray containing arr[i](starting from any...
// sum subarray between current // starting and ending points for(intk = i; k <= j; k++) result += arr[k] ; } } returnresult ; } If we take a close look then we observe a pattern. Let take an example arr[] = [1, 2, 3], n = 3 All subarrays : [1], [1, 2], ...
3011. 判断一个数组是否可以变为有序 Find if Array Can Be Sorted 力扣 LeetCode 题解 07:02 2974. 最小数字游戏 Minimum Number Game 力扣 LeetCode 题解 02:29 2972. 统计移除递增子数组的数目 II Count the Number of Incremovable Subarrays 力扣 LeetCode 题解 12:00 2970. 统计移除递增子数组...
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. Return the number of sub-arrays with odd sum. As the answer may grow large, the answer must be computed modulo 10^9 + 7. Example 1: Input: arr = [1,3,5] Output: 4 Explanation: All sub-arrays are [[1],[1,3],[1,3,5],[3],[3,5],[5]]...
1588. Sum of All Odd Length Subarrays Given an array consisting ofnintegers, find the contiguous subarray of given lengthkthat has the maximum average value. And you need to output the maximum average value. Example 1: Input: [1,12,-5,-6,50,3], k = 4...
Given an \\\(n imes n\\\) array A of integers, with at least one positive value, the maximum subarray sum problem consists in finding the maximum sum among the sums of all rectangular subarrays of A. The maximum subarray problem appears in several scientific applications, particularly in ...
A Subarray is an array that is the contiguous part of an array. Consider the given arraynums;we have to find the contiguous array containing positive and negative numbers, returns its sum. Suppose the array is: A: [-2, 1, -3, -1, 2, 1, -5, 4] ...
思路:A[i]是最低值的次数 = (左边连续的大于A[i]的个数+1)*(右边连续大于等于A[i]的个数+1) 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])...
classSolution{publicint[]maxSumOfThreeSubarrays(int[]nums,intk){intn=nums.length;int[]sums=newint[n];//sum of nums[0] to nums[i]int[]left=newint[n];int[]right=newint[n];sums[0]=nums[0];for(inti=1;i<n;i++){sums[i]=sums[i-1]+nums[i];}intmax=Integer.MIN_VALUE;//...