[ACCEPTED]-Words combinations without repetition-words

What you are trying to do is get all the 2 permutations of a collection.

• Unique permutations of list
• permutations of k objects from a set of n algorithm

Here is the 1 code snippet:

static void Main(string[] args)
{
    var list = new List<string> { "a", "b", "c", "d", "e" };
    var result = GetPermutations(list, 3);
    foreach (var perm in result)
    {
        foreach (var c in perm)
        {
            Console.Write(c + " ");
        }
        Console.WriteLine();
    }
    Console.ReadKey();
}

static IEnumerable<IEnumerable<T>> GetPermutations<T>(IEnumerable<T> items, int count)
{
    int i = 0;
    foreach (var item in items)
    {
        if (count == 1)
            yield return new T[] { item };
        else
        {
            foreach (var result in GetPermutations(items.Skip(i + 1), count - 1))
                yield return new T[] { item }.Concat(result);
        }

        ++i;
    }
}

Outputs:

a b c 
a b d 
a b e 
a c d 
a c e 
a d e 
b c d 
b c e 
b d e 
c d e 

Here's what I put together:

static class LinqExtensions
{
    public static IEnumerable<IEnumerable<T>> CombinationsWithoutRepetition<T>(
        this IEnumerable<T> items,
        int ofLength)
    {
        return (ofLength == 1) ?
            items.Select(item => new[] { item }) :
            items.SelectMany((item, i) => items.Skip(i + 1)
                                               .CombinationsWithoutRepetition(ofLength - 1)
                                               .Select(result => new T[] { item }.Concat(result)));
    }

    public static IEnumerable<IEnumerable<T>> CombinationsWithoutRepetition<T>(
        this IEnumerable<T> items,
        int ofLength,
        int upToLength)
    {
        return Enumerable.Range(ofLength, Math.Max(0, upToLength - ofLength + 1))
                         .SelectMany(len => items.CombinationsWithoutRepetition(ofLength: len));
    }

}

...

foreach (var c in new[] {"a","b","c","d"}.CombinationsWithoutRepetition(ofLength: 2, upToLength: 4))
{
    Console.WriteLine(string.Join(',', c));
}

produces:

a,b
a,c
a,d
b,c
b,d
c,d
a,b,c
a,b,d
a,c,d
b,c,d
a,b,c,d

Note 11 that this is concise but inefficient and 10 should not be used for large sets or inner 9 loops. Notably, the short arrays are re-created 8 multiple times and could be memoized, and 7 the IEnumerable will be iterated multiple times, which 6 can cause unexpected work if care is not 5 taken.

Also, if the input contains duplicates 4 then the output will as well. Either use 3 .Distinct().ToArray() first, or use another solution which includes 2 equality checking and, presumably, takes 1 an IEqualityComparer for generality.

public IActionResult Index() 
{
    var list = new List<string> { "a", "b", "c", "d", "e" };
    List<string> ret = GetAllCombinations(list).OrderBy(_ => _).ToList();

    return View();
}

static IEnumerable<string> GetAllCombinations(IEnumerable<string> list)
{
    return list.SelectMany((mainItem, mi) => list.Where((otherItem, oi) => mi < oi)
                              .Select(otherItem => mainItem + otherItem));
}

Ouput ret:

ab
ac
ad
ae
bc
bd
be
cd
ce
de

0

What about a more functional solution

var list = new List<string> { "a", "b", "c", "d", "e" };
GetAllCombinations(list).OrderBy(_ => _).ToList().ForEach(Console.WriteLine);


static IEnumerable<string> GetAllCombinations(IEnumerable<string> list)
{
    return list.SelectMany(mainItem => list.Where(otherItem => !otherItem.Equals(mainItem))
                              .Select(otherItem => mainItem + otherItem));
}

Ouput:

ab
ac
ad
ae
ba
bc
bd
be
ca
cb
cd
ce
da
db
dc
de
ea
eb
ec
ed

0