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]) ...
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], [...
示例 1: 输入:arr = [3,1,2,4] 输出:17 解释: 子数组为 [3],[1],[2],[4],[3,1],[1,2],[2,4],[3,1,2],[1,2,4],[3,1,2,4]。&n...[LeetCode] 907. Sum of Subarray Minimums 子数组最小值之和 Given an array of integers A, find the sum of min(B), where B ...
Sum of Subarray Minimums 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......
Given a circular array C of integers represented by A, find the maximum possible sum of a non-empty subarray of C. Here, a circular array means the end of the array connects to the beginning of the array. (Formally, C[i] = A[i] when 0 <= i < A.length, and C[i+A.length] ...
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...
1classSolution {2func sumSubarrayMins(_ A: [Int]) ->Int {3ifA.isEmpty {4return05}67let n = A.count, MOD =10000000078vars =[Int]()9//sum of subarray mins ending with A[i]10vardp = [Int](repeating:0, count: n)11//3 1 2 4 5 112foriin0..<n {13dp[i] = (dp[i] +...
minimum in O(1) time, we'll get the total sum in O(n) time. To do this, we need to find out the left and right effect bound of A[i] as the minimum in subarrays. If we have these two bounds, then A[i]'s contribution to the total sum is just A[i] * (i - leftBound)...
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