for just max sum subarray it is Kadane's algorithm in O(n) time, however, I couldn't think of a way to solve for atleast 2 numbers faster than O(n^2). Any idea or a solution?
这个问题是这样的(https://leetcode.com/problems/maximum-subarray/): Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum. 就是给一组整数,找到一组连续的子序列,让子序列的和最大,并且返回最大和。 比如说: 输入...
To find the combination of continuous ranges that returns the maximum sum, you can use the Kadane...
Given an array, find the maximum subset sum of that array. A subset is a set of consecutive elements in the array. Input The first line of the input contains an integer t, denoting the number of test cases. Each test case contains 2 lines. The first line consists of a number n, the...
以及这道题目:max subarray sum no more than k 本题基本上是以上两种思路的组合了,具体的实现很简单。 题目描述 Given a non-empty 2D matrix matrix and an integer k, find the max sum of a rectangle in the matrix such that its sum is no larger than k. ...
// you cannot add this in this subarray, make new one // say you add this num in new subarray, then sum = num sum = num; pieces++; } else { sum += num; } }if (pieces > m) { start = mid + 1; } // else if (pieces <= m) { end = mid }; ...
We first loop through the input to build S. Afterwards, we do what they suggest in the text. This is n+n=2n=Θ(n).typedef struct { unsigned left; unsigned right; int sum; } max_subarray; max_subarray find_maximum_subarray(int
First thing to note is that sum of subarray (i,j] is just the sum of the first j elements less the sum of the first i elements. Store these cumulative sums in the array cum. Then the problem reduces to finding i,j such that i<j and cum[j]−cum[i] is as close to k but ...
For each test case, you should output a line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the minimum start position, if still more than one , output the ...
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