Divide an integer into groups with C#

Say you’d like to divide an integer, e.g. 20, into equal parts of 3 and distribute any remainder equally across the groups. The result of such an operation would be the following 3 integers:

7,7,6

20 can be divided into 3 equal parts of 6 and we have a remainder of 20 – 6 * 3 = 2. 2 is then added as 1 and 1 to the first two groups of 6. The result is a more or less equal distribution of the start integer.

The following function will perform just that:

public static IEnumerable<int> DistributeInteger(int total, int divider)
{
	if (divider == 0)
	{
		yield return 0;
	}
	else
	{
		int rest = total % divider;
		double result = total / (double)divider;

		for (int i = 0; i < divider; i++)
		{
			if (rest-- > 0)
				yield return (int)Math.Ceiling(result);
			else
				yield return (int)Math.Floor(result);
		}
	}
}

Call it as follows:

List<int> test = DistributeInteger(20, 3).ToList();

“test” will include 7,7,6 as expected.

View all various C# language feature related posts here.