# Find all available combinations

I have one done the same with strings. However, I just swapped random digits until I get all combinations(which is the factorial of the length of the string). I then print out the non-recurring ones.

Here is how I did it, just change the String you input to the allpos() method and it will work for any number.

``````import java.util.Random;
import java.util.Scanner;

public class shift {

public static void main(String[] args) {

//Change "012" to anything, even a String.
allpos("012");
}
public static int unbin(String s){

int ans=0;
for(int x=0;x<s.length();x++){
char c=s.charAt(x);
if(c=='1'){
ans+=Math.pow(2, s.length()-1-x);
}

}
return ans;

}
public static String bin(int n){
int[] ans=new int[32];
int a=n;
for(int x=31;x>=0;x--){
if(Math.pow(2, x)<=a){
a-=Math.pow(2, x);
ans[x]=1;
}
}
String s="";boolean zero=false;
for(int x=ans.length-1;x>=0;x--){
if(ans[x]==1){
s+="1";
zero=true;
continue;
}else if(zero){
s+="0";
}
}
return s;
}
public static boolean ispresent(String[] words,String s){
for(int x=0;x<words.length;x++){
if(words[x]==null){return false;}
if(words[x].equals(s)){return true;}

}
return false;

}
public static void allpos(String s){

int fac=1;
for(int x=1;x<=s.length();x++){
fac*=x;
}
int len=s.length();
String[] values=new String[fac];
values[0]=s;
System.out.println(s+" (1)");
int id1=0,id2=0;
Random r=new Random();
for(int x=1;x<fac;){

s=inchange(s,r.nextInt(len),r.nextInt(len));
if(ispresent(values,s)==false){
values[x]=s;
x++;
System.out.println(s+" ("+x+")");
}

}

}
public static String inchange(String s,int id1,int id2){
char[] chars=new char[s.length()];
for(int x=0;x<s.length();x++){
chars[x]=s.charAt(x);

}
char temp=chars[id1];
chars[id1]=chars[id2];
chars[id2]=temp;
String ans="";
for(int x=0;x<s.length();x++){
ans+=chars[x];

}
return ans;
}

}
``````

I know that it is inefficient but it gets the job done for smaller numbers. I hope this helped.