Iterate all possible variants of a sequence - algorithm

Question

I have a sequence of n letters: e.g. A B C A D , or A B F A
What I want is to get every possible variant with a comma between the letters.
i.e. for A B F A
A B F A
A,B F A
A B,F A
A B F,A
A,B,F A
A B,F,A
A,B F,A
A,B,F,A
Can anyone recommend a good algorithm for doing this? Language not important.

Answer 1

Simple solution to use binary array to represent if there is a comma or not.
A B F A contains three positions where comma may be (AB, BF, FA)
That means if you create 3-element array and try every possible combination of 0 and 1 you'll get the desired result. 000, 001, 010, 011, 100, 101, 110, 111
Simple program in java prints all binary permutation for n bits:
String s = "ABFA";
int bits = s.length() - 1;
int lastNumber = (int)Math.pow(2, bits);
System.out.println(lastNumber);
for (int i = 0; i < lastNumber; i++) {
System.out.println(completeZeros(Integer.toString(i, 2), bits));
}
static String completeZeros(String s, int bits) {
String result = s;
for (int i = 0; i < bits - s.length(); i++) {
result = "0" + result;
}
return result;
}
To apply binary permutation "010" to string "ABFA" use next function:
static String applyBinary(String s, String binary) {
String result = "" + s.charAt(0);
for (int i = 0; i < binary.length(); i++) {
if (binary.charAt(i) == '1') result += ", ";
result += s.charAt(i + 1);
}
return result;
}
The output is:
ABFA
ABF, A
AB, FA
AB, F, A
A, BFA
A, BF, A
A, B, FA
A, B, F, A

Answer 2

Use binary system for this task.
1 means comma is present, 0 means comma is not present. Each position in number informs about presence of another comma. For example for AFA:
00 : A F A
01 : A F,A
10 : A,F A
11 : A,F,A
Numbers must be taken from range [0 .. (n-1)^2-1]

Answer 3

Here's a simple JavaScript demonstration.
var str = "ABFA";
function algo(str) {
var result = [];
var n = str.length;
var total = Math.pow(n-1, 2) - 1;
for(var mask = 0; mask < total; mask++) {
var bits = mask;
var newstr = "";
for(var i=0; i<str.length - 1; i++, bits>>>=1) {
var hasComma = (bits & 1) == 1;
newstr += str.charAt(i);
newstr += (hasComma ? "," : " ");
}
newstr += str.charAt(str.length - 1);
result.push(newstr);
}
return result;
}
algo(str);
You calculate the total number of combinations "total"
You count up to that number "mask"
You use the binary bits of your counter "bits" to add commas

Answer 4

There are two approaches to this problem.
1.Recursive ( Start = "printAllComb ( s , "" , 0 );" )
printAllComb(string s, string const, int i)
{
if ( i == s.length() )
print const
printAllComb(s,const+string.at(i)+',',i+1);
printAllComb(s,const+string.at(i),i+1);
}
2.Dynamic Programming
char binaryS[s.length]="0000";
//Basically no. of zeros = no. of Alphabets in the string
//define a function AddOne() which adds 1 to the character representation
//AddOne() modifies the character array such that it stays in the bit representation
//Characters because to save space
while ( All the bits are not one )
{
for ( int i=0; i<s.length(); i++ )
{
print s.at(i)
if ( binaryS.at(i) == '1' )
print ","
}
print "\n"
AddOne();
}