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I have an arraylist of strings "abcde" I want to a method to return another arraylist with all the possible combination of a given arraylist (ex:ab,ac,ad...) in C#

anyone knows a simple method?

NB: all possible combinations of length 2, and would be better if the length is variable(can be changed)

Mona
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3 Answers3

10

Pertaining your comment requiring combinations of length two:

string s = "abcde";
var combinations = from c in s
                   from d in s.Remove(s.IndexOf(c), 1)
                   select new string(new[] { c, d });
foreach (var combination in combinations) {
    Console.WriteLine(combination);
}

Responding to your edit for any length:

static IEnumerable<string> GetCombinations(string s, int length) {
    Guard.Against<ArgumentNullException>(s == null);
    if (length > s.Length || length == 0) {
        return new[] { String.Empty };
    if (length == 1) {
        return s.Select(c => new string(new[] { c }));
    }
    return from c in s
           from combination in GetCombinations(
               s.Remove(s.IndexOf(c), 1),
               length - 1
           )
           select c + combination;
}

Usage:

string s = "abcde";
var combinations = GetCombinations(s, 3);
Console.WriteLine(String.Join(", ", combinations));

Output:

abc, abd, abe, acb, acd, ace, adb, adc, ade, aeb, aec, aed, bac, bad, bae, bca,
bcd, bce, bda, bdc, bde, bea, bec, bed, cab, cad, cae, cba, cbd, cbe, cda, cdb,
cde, cea, ceb, ced, dab, dac, dae, dba, dbc, dbe, dca, dcb, dce, dea, deb, dec,
eab, eac, ead, eba, ebc, ebd, eca, ecb, ecd, eda, edb, edc
jason
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  • Also you should do var retComb = GetCombinations(s,2).Distinct().Where(p=>p[0] >= p[p.Length]); to remove redundant items, and reverse of items, for example for `aabd` test your approach – Saeed Amiri Nov 14 '10 at 10:25
5

Here is my generic function which can return all the combinations of type T:

static IEnumerable<IEnumerable<T>> GetCombinations<T>(IEnumerable<T> list, int length)
{
    if (length == 1) return list.Select(t => new T[] { t });

    return GetCombinations(list, length - 1)
        .SelectMany(t => list, (t1, t2) => t1.Concat(new T[] { t2 }));
}

Usage:

Console.WriteLine(
    string.Join(", ",
        GetCombinations("abcde".ToCharArray(), 2).Select(list => string.Join("", list))
    )
);

Output:

aa, ab, ac, ad, ae, ba, bb, bc, bd, be, ca, cb, cc, cd, ce, da, db, dc, dd, de, ea, eb, ec, ed, ee

UPDATED Please see my answer here for other scenarios, e.g. permutations and k-combinations etc.

Community
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Pengyang
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0

Combination of numbers in array by using arrays only and recursion:

    static int n = 4;
    int[] baseArr = { 1, 2, 3, 4 };
    int[] LockNums;

    static void Main(string[] args)
    {
        int len = baseArr.Length;
        LockNums = new int[n];

        for (int i = 0; i < n; i++)
        {
            int num = baseArr[i];
            DoCombinations(num, baseArr, len);
            //for more than 4 numbers the print screen is too long if we need to check the result next line will help
            //Console.ReadLine(); 

        }
    }

    private void DoCombinations(int lockNum, int[] arr, int arrLen )        
    {
        int n1 = arr.Length;
        // next line shows the difference in length between the previous and its previous array
        int point = arrLen - n1; 
        LockNums[n - arr.Length] = lockNum;

        int[] tempArr = new int[arr.Length - 1];
        FillTempArr(lockNum, arr, tempArr);

        //next condition will print the last number from the current combination
        if (arr.Length == 1)
        {
            Console.Write(" {0}", lockNum);
            Console.WriteLine();
        }

        for (int i = 0; i < tempArr.Length; i++)
        {
            if ((point == 1) && (i != 0))
            {
                //without this code the program will fail to print the leading number of the next combination
                //and 'point' is the exact moment when this code has to be executed
                PrintFirstNums(baseArr.Length - n1);
            }
            Console.Write(" {0}", lockNum);
            int num1 = tempArr[i];
            DoCombinations(num1, tempArr, n1);
        }
    }

    private void PrintFirstNums(int missNums)
    {
        for (int i = 0; i < missNums; i++)
        {
            Console.Write(" {0}", LockNums[i]);
        }
    }

    private void FillTempArr(int lockN, int[] arr, int[] tempArr)
    {
        int idx = 0;
        foreach (int number in arr)
        {
            if (number != lockN)
            {
                tempArr[idx++] = number;
            }
        }
    }

    private void PrintResult(int[] arr)
    {
        foreach (int num in arr)
        {
            Console.Write(" {0}", num);
        }
    }
Kiril Dobrev
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  • Hello, this is my solution for a combination of numbers. The solution uses only arrays. I'm doing this post because I have not met a solution with arrays anywhere. The code is set for a quick demonstration, otherwise it should work with any array length. – Kiril Dobrev Oct 29 '17 at 10:55