It is possible to generate non-overlapping sublists of length n of a given list
For example, if a list is (A, B, B, C), then the non-overlapping sublists of length 2 will be:
To do that you should use an instance of the complex combination generator
// create a vector (A, B, B, C) ICombinatoricsVector<String> vector = Factory.createVector(new String[] { "A", "B", "B", "C" }); // Create a complex-combination generator Generator<ICombinatoricsVector<String>> gen = new ComplexCombinationGenerator<String>(vector, 2); // Iterate the combinations for (ICombinatoricsVector<ICombinatoricsVector<String>> comb : gen) { System.out.println(ComplexCombinationGenerator.convert2String(comb) + " - " + comb); }
And the result
([A],[B, B, C]) - CombinatoricsVector=([CombinatoricsVector=([A], size=1), CombinatoricsVector=([B, B, C], size=3)], size=2) ([B, B, C],[A]) - CombinatoricsVector=([CombinatoricsVector=([B, B, C], size=3), CombinatoricsVector=([A], size=1)], size=2) ([B],[A, B, C]) - CombinatoricsVector=([CombinatoricsVector=([B], size=1), CombinatoricsVector=([A, B, C], size=3)], size=2) ([A, B, C],[B]) - CombinatoricsVector=([CombinatoricsVector=([A, B, C], size=3), CombinatoricsVector=([B], size=1)], size=2) ([A, B],[B, C]) - CombinatoricsVector=([CombinatoricsVector=([A, B], size=2), CombinatoricsVector=([B, C], size=2)], size=2) ([B, C],[A, B]) - CombinatoricsVector=([CombinatoricsVector=([B, C], size=2), CombinatoricsVector=([A, B], size=2)], size=2) ([B, B],[A, C]) - CombinatoricsVector=([CombinatoricsVector=([B, B], size=2), CombinatoricsVector=([A, C], size=2)], size=2) ([A, C],[B, B]) - CombinatoricsVector=([CombinatoricsVector=([A, C], size=2), CombinatoricsVector=([B, B], size=2)], size=2) ([A, B, B],[C]) - CombinatoricsVector=([CombinatoricsVector=([A, B, B], size=3), CombinatoricsVector=([C], size=1)], size=2) ([C],[A, B, B]) - CombinatoricsVector=([CombinatoricsVector=([C], size=1), CombinatoricsVector=([A, B, B], size=3)], size=2)