Question

I have an object that contains two arrays, the first is a slope array:

double[] Slopes = new double[capacity];

The next is an array containing the counts of various slopes:

int[] Counts = new int[capacity];

The arrays are related, in that when I add a slope to the object, if the last element entered in the slope array is the same slope as the new item, instead of adding it as a new element the count gets incremented.

i.e. If I have slopes 15 15 15 12 4 15 15, I get:

Slopes = { 15, 12, 4, 15 }
Counts = {  3,  1, 1,  2 }

Is there a better way of finding the i_th item in slopes than iterating over the Counts with the index and finding the corresponding index in Slopes?

edit: Not sure if maybe my question wasn’t clear. I need to be able to access the i_th Slope that occurred, so from the example the zero indexed i = 3 slope that occurs is 12, the question is whether a more efficient solution exists for finding the corresponding slope in the new structure.

Maybe this will help better understand the question: here is how I get the i_th element now:

public double GetSlope(int index)
        int countIndex = 0;
        int countAccum = 0;
        foreach (int count in Counts)
        {
            countAccum += count;
            if (index - countAccum < 0)
            {
                return Slopes[countIndex];
            }
            else
            {
                countIndex++;
            }
        }
        return Slopes[Index];
}

I am wondering if there is a more efficient way?

Answer 1

You could use a third array in order to store the first index of a repeated slope

double[] Slopes = new double[capacity];
int[] Counts = new int[capacity]; 
int[] Indexes = new int[capacity]; 

With

Slopes  = { 15, 12, 4, 15 }
Counts  = {  3,  1, 1,  2 } 
Indexes = {  0,  3, 4,  5 } 

Now you can apply a binary search in Indexes to serach for an index which is less or equal to the one you are looking for.

Instead of having an O(n) search performance, you have now O(log(n)).