Question

given a string array

String[] values = {"A", "B", "C", "D", "E"};

and a constant

 final String X = "X";

Generate every possible way the value array can contain 0 or more X

i.e

    [A, B, C, D, E]
    [A, X, X, D, E]
    [A, B, X, D, E]
    [X, X, X, X, X]

and so on with every possibility. Needs to be fast – bit of a brain teaser i’m having trouble making efficient. Not as easy as it sounds.

Answer 1

Answer was in Joachim Sauer’s hint:

      String[] values = {"A", "B", "C", "D", "E"};
    final String ONE = "1";
    final String X = "X";
    int control=0;
    StringBuilder sb = new StringBuilder(values.length);
    String copy[];
    int count = Integer.parseInt(new String(new char[values.length]).replace("\0", ONE), 2) ;
    while (control <= count) {
        sb.delete(0, sb.length());
        sb.append(String.format("%" + values.length  + "s", Integer.toBinaryString(control++)).replace(' ', '0'));
        copy = values.clone();
        for (int x = 0; x < sb.length(); x ++) {
            if (sb.substring(x, x+1).equals(ONE)) {
                copy[x] = X;
            }
        }
       System.out.println(control + " " + Arrays.toString(copy));
       map.put(generateHashCode(copy),copy.clone());
    }


    System.out.println(map.values().size());