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

Question

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 1

The simplest algorithm for generating subsets of a set of size N is to consider all binary numbers using N bits. Each position in the number represents an element from the set. If a bit in the number is 1, the corresponding set element is in the subset, otherwise the element isn't in the subset. Since the bits in a number are ordered, this preserves the ordering of the original set.

References:

1. "Efficiently Enumerating the Subsets of a Set"; Loughry, Hemert and Schoofs
2. "Generating Subsets"; Stony Brook Algorithm Repository

Answer 2

In C++ given the following routine:

template <typename Iterator>
bool next_combination(const Iterator first, Iterator k, const Iterator last)
{
   /* Credits: Mark Nelson http://marknelson.us */
   if ((first == last) || (first == k) || (last == k))
      return false;
   Iterator i1 = first;
   Iterator i2 = last;
   ++i1;
   if (last == i1)
      return false;
   i1 = last;
   --i1;
   i1 = k;
   --i2;
   while (first != i1)
   {
      if (*--i1 < *i2)
      {
         Iterator j = k;
         while (!(*i1 < *j)) ++j;
         std::iter_swap(i1,j);
         ++i1;
         ++j;
         i2 = k;
         std::rotate(i1,j,last);
         while (last != j)
         {
            ++j;
            ++i2;
         }
         std::rotate(k,i2,last);
         return true;
      }
   }
   std::rotate(first,k,last);
   return false;
}

You can then proceed to do the following:

std::string s = "12345";
for(std::size_t i = 1; i <= s.size(); ++i)
{
   do
   {
      std::cout << std::string(s.begin(),s.begin() + i) << std::endl;
   }
   while(next_combination(s.begin(),s.begin() + i,s.end()));
}

Answer 3

using python, the itertools module defines a combinations() method which does just what you need.

from itertools import *
list(combinations( '12345', 2 ))

will give you:

[('1', '2'), ('1', '3'), ('1', '4'), ('1', '5'), ('2', '3'), ('2', '4'), ('2', '5'), ('3', '4'), ('3', '5'), ('4', '5')]

Answer 4

Adrien Plisson's answer shows how one retrieves all subsequences of a specified length in Python (for arbitrary sequence data types). The OP specifies that he works with strings, and that he wants all subsequences. Thus, using itertools.combinations we define:

>>> from itertools import combinations
>>> def subseq_combos(inp):
...     return (''.join(s) for r in range(len(inp) + 1) for s in combinations(inp, r))
... 
>>> list(subseq_combos('12345'))
['', '1', '2', '3', '4', '5', '12', '13', '14', '15', '23', '24', '25', '34', '35', '45', '123', '124', '125', '134', '135', '145', '234', '235', '245', '345', '1234', '1235', '1245', '1345', '2345', '12345']

(If the empty subsequence should be omitted, then use range(1, len(inp) + 1)).)

Answer 5

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;
        }
    }

Answer 6

You can use the following class for this (in Java):

class Combinations {

  String input;
  StringBuilder cur;

  private void next(int pos, int reminder) {
    cur.append(input.charAt(pos));

    if (reminder == 1) {
      System.out.println(cur);
    } else {
      for (int i = pos + 1; i + reminder - 1 <= input.length(); i++)
        next(i, reminder - 1);
    }
    cur.deleteCharAt(cur.length() - 1);
  }

  public void generate(String input) {
    cur = new StringBuilder();
    this.input = input;
    for (int length = 1; length <= input.length(); length++)
      for (int pos = 0; pos + length <= input.length(); pos++)
        next(pos, length);
  }
}

To run your example use the following code:

new Combinations().generate("12345");

The order of the output is the same as in example. It does not require to store all subsets and then sort them to obtain the order you described.