Given a length and a set of characters, how to get all the possible string combinations [duplicate]

Answer 1

Think of it in the same way you would counting. You're technically "counting" from aaaa to bbbb, like binary.

aaaa -> 0000
aaab -> 0001
aaba -> 0010
aabb -> 0011
...
bbbb -> 1111

Without seeing what you've tried, I can't help you more than that, but essentially you need to enumerate all the "numbers" between your "lowest" element and your "highest" element by counting up through them.

For higher element counts, just treat your counting as counting in a higher base. For eight elements, Set = {a, b, c, d, e, f, g, h}, you would be counting up in what's essentially octal:

aaaa -> 0000
aaab -> 0001
...
aaah -> 0007
aaba -> 0010
...
hhhh -> 7777

It's the same way you enumerate all the combinations of 0-9 with a length of 4, by counting from 0000 to 9999.

Edit:

Thank you for posting your code. You're correct, you're doing permutations. A better way would be to use a multi-combination (combination with repeated elements in the ordererd combination set) algorithm like the one discussed here.

Answer 2

You've got an answer, but I feel that you need some more help:

if(string.length() == 0)
{
System.out.println(result);
}

Why would you print an empty string? It will print nothing at all. You may want to print a message and exit from your function.

Answer 3

This may be error prone, as I haven't tested it, but it should work. Even if it doesn't, it should be extremely close to what you are looking for. Note that the first for loops populate the resultList instead of setting values, as there will be nothing to append to.

Let me know if there is an issue and I will correct it.

ArrayList<String> results = new ArrayList<String>();
ArrayList<String> components = new ArrayList<String>(){"a","b","c"};
int n = 4;

int size = components.size();
for ( int j = 0; j < size; j++ )
{
  // start with size^(n-1) copies of each letter. 
  for ( int i = 0; i < Math.pow( size, n-1); i++ )
  {
    results.add( components.get( j ) );
  }
}

// At this point you have each letter in there once...

for( int depth = 1; depth < n; depth++ )
{
  for( int j = 0; j < size; j++ )
  {
    String toAppend = components.get( j );
    for( int i = j; i < results.size(); i += size )
    {
      String current = results.get( i );
      current += toAppend;
      results.set( i, current );
    }
  }
}