How to obtain all subsequence combinations of a String (in Java, or C++ etc)

Let’s say I’ve a string “12345” I should obtain all subsequence combinations of this string such as:

1. –> 1 2 3 4 5
2. –> 12 13 14 15 23 24 25 34 35 45
3. –> 123 124 125 234 235 345
4. –> 1234 1235 1245 1345 2345
5. –> 12345

Please note that I grouped them in different number of chars but not changed their order. I need a method/function does that.

Answer

You want a powerset. Here are all the questions on StackOverflow that mention powersets or power sets.

Here is a basic implementation in python:

def powerset(s):
    n = len(s)
    masks = [1<<j for j in xrange(n)]
    for i in xrange(2**n):
        yield [s[j] for j in range(n) if (masks[j] & i)]


if __name__ == '__main__':
    for elem in powerset([1,2,3,4,5]):
        print elem

And here is its output:

[]
[1]
[2]
[1, 2]
[3]
[1, 3]
[2, 3]
[1, 2, 3]
[4]
[1, 4]
[2, 4]
[1, 2, 4]
[3, 4]
[1, 3, 4]
[2, 3, 4]
[1, 2, 3, 4]
[5]
[1, 5]
[2, 5]
[1, 2, 5]
[3, 5]
[1, 3, 5]
[2, 3, 5]
[1, 2, 3, 5]
[4, 5]
[1, 4, 5]
[2, 4, 5]
[1, 2, 4, 5]
[3, 4, 5]
[1, 3, 4, 5]
[2, 3, 4, 5]
[1, 2, 3, 4, 5]

Notice that its first result is the empty set. Change the iteration from this for i in xrange(2**n): to this for i in xrange(1, 2**n): if you want to skip an empty set.

Here is the code adapted to produce string output:

def powerset(s):
    n = len(s)
    masks = [1<<j for j in xrange(n)]
    for i in xrange(2**n):
        yield "".join([str(s[j]) for j in range(n) if (masks[j] & i)])

---

Edit 2009-10-24

Okay, I see you are partial to an implementation in Java. I don’t know Java, so I’ll meet you halfway and give you code in C#:

    static public IEnumerable<IList<T>> powerset<T>(IList<T> s)
    {
        int n = s.Count;
        int[] masks = new int[n];
        for (int i = 0; i < n; i++)
            masks[i] = (1 << i);
        for (int i = 0; i < (1 << n); i++)
        {
            List<T> newList = new List<T>(n);
            for (int j = 0; j < n; j++)
                if ((masks[j] & i) != 0)
                    newList.Add(s[j]);
            yield return newList;
        }
    }