Sum and split array like below.

int arr[]={1,2,3,4,5,6,7,8,9,9,8,7,6,5,4,3,2,1,9,8,7,6,5,4,3,2,1,1,2,3,4,5,6,7,8,9}

int innergroup=3;//vary from 1,2,..n

int outergroup=9; //vary from 1,2,..n

int mplit=2; //vary from 1,2,..n

res[0]=29 //1+4+7+9+6+3

res[1]=30 //2+5+8+8+5+2

res[2]=30 //3+6+9+7+4+1

res[3]=30 //9+6+3+1+4+7

res[4]=30 //8+5+2+2+5+8

res[5]=30 //7+4+1+3+6+9

res[0]=first element of inner ingroup1 of split1+first element of inner ingroup2 of split1+first element of inner ingroup3 of split1+first element of inner ingroup1 of split2+first element of inner ingroup2 of split2+first element of inner ingroup3 of split2

and likewise

class Program
{
static void Main(string[] args)
{
SplitTheArrayAndDoSumOfEveryOtherThirdNumber();
}
private static void SplitTheArrayAndDoSumOfEveryOtherThirdNumber(int split = 4)
{
//int[] arr = { 1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 4 };
int[] arr ={
1,2,3,4,5,6,7,8,9
,9,8,7,6,5,4,3,2,1
,9,8,7,6,5,4,3,2,1
,1,2,3,4,5,6,7,8,9
};
int outergroup = 9;
int factor = arr.Length / outergroup;
Console.WriteLine(arr.Length);
//int split = 2;
List<int[]> splitted = new List<int[]>();
for (int i = 0; i < split; i++)
{
splitted.Add(new int[outergroup]);
for (int j = i * outergroup; j < (i * outergroup) + (outergroup); j++)
{
(splitted[i])[j - (i * outergroup)] = arr[j];
}
Console.WriteLine();
}
List<int> sums = new List<int>();
int innergroup = 3;
for (int i = 0; i < 3; i+=2)
{
for (int y = 0; y < innergroup; y++)
{
int sum = 0;
for (int j = y; j < outergroup; j+=3)
{
sum += (splitted[i])[j] + (splitted[i + 1])[j];
}
sums.Add(sum);
}
}
foreach (int sum in sums)
{
Console.WriteLine(sum);
}
}
}

1 Reply

Know the answer? Post it — somebody with the same question will find it here.

Tapan Patel9y ago
1. class Program
2. {
3. static void Main(string[] args)
4. {
5. SplitTheArrayAndDoSumOfEveryOtherThirdNumber();
6. }
7. private static void SplitTheArrayAndDoSumOfEveryOtherThirdNumber(int split = 4)
8. {
9. //int[] arr = { 1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 4 };
10. int[] arr ={
11. 1,2,3,4,5,6,7,8,9
12. ,9,8,7,6,5,4,3,2,1
13. ,9,8,7,6,5,4,3,2,1
14. ,1,2,3,4,5,6,7,8,9
15. };
16. int outergroup = 9;
17. int factor = arr.Length / outergroup;
18. Console.WriteLine(arr.Length);
19. //int split = 2;
20. List<int[]> splitted = new List<int[]>();
22. for (int i = 0; i < split; i++)
23. {
24. splitted.Add(new int[outergroup]);
25. for (int j = i * outergroup; j < (i * outergroup) + (outergroup); j++)
26. {
27. (splitted[i])[j - (i * outergroup)] = arr[j];
28. }
29. Console.WriteLine();
30. }
31. List<int> sums = new List<int>();
32. int innergroup = 3;
33. for (int i = 0; i < 3; i+=2)
34. {
36. for (int y = 0; y < innergroup; y++)
37. {
38. int sum = 0;
39. for (int j = y; j < outergroup; j+=3)
40. {
41. sum += (splitted[i])[j] + (splitted[i + 1])[j];
42. }
43. sums.Add(sum);
44. }
47. }
49. foreach (int sum in sums)
50. {
51. Console.WriteLine(sum);
52. }
53. }
54. }
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