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:


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;
		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);
				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.


About Andras Nemes
I'm a .NET/Java developer living and working in Stockholm, Sweden.

2 Responses to Divide an integer into groups with C#

  1. Pingback: Divide an integer into groups with C# | .NET developpement

  2. vicky says:

    please give code in VB.NET for distributing the integer in equal parts

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