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?
// Package maxsubarraysum is a package containing a solution to a common // problem of finding max contiguous sum within a array of ints. package maxsubarraysum import ( "github.com/TheAlgorithms/Go/math/max" ) // MaxSubarraySum returns the maximum subarray sum func MaxSubarraySum(array ...
1publicintmaxSubArray(int[] A) {2intnewsum=A[0];3intmax=A[0];4for(inti=1;i<A.length;i++){5newsum=Math.max(newsum+A[i],A[i]);6max=Math.max(max, newsum);7}8returnmax;9} 1intmaxSubArray(int*a,constintlength)2{3intmaxSumSubArray =0;4intsum_i =0;5for(inti=0; i ...
MaxSubArraySum是一种经典的算法问题,其目标是在给定的整数数组中找到具有最大总和的连续子数组。解决这个问题的常见方法是使用动态规划。算法的基本思想是维护两个变量:当前子数组的最大和以及全局最大和。在遍历数组时,我们不断更新当前子数组的最大和,如果该和变得小于0,则重新开始计算子数组和,并更新全局最大和...
1#include <STDIO.H>2typedefstructSU_tag{3SU_tag(){}4SU_tag(inta,intb,intc):max_sum(a),left(b),right(c){}5intmax_sum;6intleft;7intright;8}SU;910SU find_max_crossing_subarray(int*a,intlow,intmid,inthigh)11{12intleft,left_max=a[mid],right,right_max=a[mid+1],i,sum;13su...
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 ...
bemin(Ai,Ai+1,...,Aj)∗max(Ai,Ai+1,...,Aj)\text{min}(A_i, A_{i+1},...,A_j) * \text{max}(A_i, A_{i+1},...,A_j)min(Ai,Ai+1,...,Aj)∗max(Ai,Ai+1,...,Aj). The costGGGof a partition is the sum ofFFFfor all its subarrays....
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++)...
Kadane's algorithm. Kadane's algorithm is an efficient way to find the maximum subarray sum in ...
这题被标记为hard,第一眼看到,确实很容易想到是dp,但思路之后就陷入混乱,原因就是不知道dp的状态转移方程,以及dp过程的开始和结束,本题事实上是两道题目的组合,分别是 Max Sum of Rectangle in a matrix,这道题的题解可以看这个视频(需要翻墙) 以及这道题目:max subarray sum no more than k 本题基本上...