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xellez
xellez12/8/2022
Wrong count for my inversions function

It is supposed to be 10 inversion count but I am getting 13
           var array = new int[] { 1, 8, 2, 0, 4, 3, 7, 6, 5 };

            int inv_count = 0;
            for (int i = 0; i < array.Length - 1; i++) {
                for (int j = i + 1; j < array.Length; j++)
                {
                    if (array[i] > array[j])
                        inv_count++;
                }   
            }
            return inv_count;
        } 
What am I screwing up or missing something ?
TheBoxyBear
TheBoxyBear12/8/2022
What do you mean by inversion?
xellez
xellez12/8/2022
it is for a puzzle problem
xellez
xellez12/8/2022
Now find the number of inversion, by counting tiles precedes the another tile with lower number.

In our case, 1,2,3,4,5,6,7 is having 0 inversions, and 8 is having 1 inversion as it's preceding the number 7.

Total number of inversion is 1 (odd number) so the puzzle is insolvable.
xellez
xellez12/8/2022
trying to apply the same method
Jimmahdean
Jimmahdean12/8/2022
just counting, i get 12 13
Jimmahdean
Jimmahdean12/8/2022
but i'm lacking about four cups of coffee so idk
Jimmahdean
Jimmahdean12/8/2022
i'm not sure what you're trying to do really
Jimmahdean
Jimmahdean12/8/2022
1 precedes 0 : 1
8 precedes 2 0 4 3 7 6 5 : 7 (8)
2 precedes 0 : 1 (9)
0 precedes nothing
4 precedes 3 : 1 (10)
3 precedes nothing
7 precedes 6 5 : 2 (12)
6 precedes 5 : 1 (13)
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