我最近遇到了一个算法问题，如下所示。
有两个大小为'N'的数组。 找到选择元素子集的算法，使得Array1之和与Array2之和之间的绝对差值应为＆lt; ='M'（其中M为常数）。
Eg:
Array1 -> 1,2
Array2 -> 4,5
M =2
Total number of subsets are '6'
The subsets are like below
( sum of array1, sum of array2 )
(0,0) -> no numbers selected
(1,0) -> selecting 1 from array1, selecting none from array2
(2,0) -> selecting 2 from array1, selecting none from array2
(2,4) -> selecting 2 from array1, selecting 4 from array2
(3,4) -> selecting 1,2 from array1, selecting 4 from array2
(3,5) -> selecting 1,2 from array1, selecting 5 from array2
算法的复杂性应为O（N * M）
提前致谢