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? UPD: Thanks everyone. Now I know how
First thing to note is that sum of subarray(i,j]is just the sum of the firstjelements less the sum of the firsti elements. Store these cumulative sums in the array cum. Then the problem reduces to findingi,jsuch thati<j andcum[j]−cum[i] is as close tok but lower than it. ...
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 m...
We define the cost FF of a subarray A[i,j]A[i,j] to be min(Ai,Ai+1,...,Aj)∗max(Ai,Ai+1,...,Aj)min(Ai,Ai+1,...,Aj)∗max(Ai,Ai+1,...,Aj). The cost GG of a partition is the sum of FF for all its subarrays. Print the sum of GG for all 2N−12N−...
Also, subset of consecutive elements is probably better defined as subarray. avidcoder: 2017-08-15 09:11:39 Weak test cases! Since it is a tutorial problem. It's okay. Last edit: 2017-08-15 09:33:11 Francky: 2014-12-28 16:33:34 Description gives : "All input fits in the ...
Kadane's algorithm is an efficient way to find the maximum subarray sum in an array of numbers...
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Runtime:172 ms, faster than55.61% of C++ online submissions for Max Sum of Rectangle No Larger Than K. here is a esay way to understand "largest sum of contiguous subarray No Larger than k" privateintmaxSumSubArray(int[]a,intk){intmax=Integer.MIN_VALUE;for(inti=0;i<a.length;i++)...
软RAID(software-based RAID)是基于软件的RAID。它可能是最普遍的被使用的RAID阵列，这是由于现在的很多...
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