Question

I have two strings str1 and str2.
Is there any algorithm that can be used in order to print out all interleavings of the two strings using recursion?

Update:

public class Interleave {


    private String resultString[] = new String[10];
    private String[] interStr(String str1, String str2){
    int n = ((Factorial.factorial(str1.length() + str2.length())) / (Factorial.factorial(str1.length()) * Factorial.factorial(str2.length())));
    //n is number of interleavings based on (str1.length()+str2.length())! / (str1.length()! * str2.length()!)
    if(str1.length() == 0){
        resultString[0] = str2;
        return resultString;
    }

    if(str2.length() == 0){
        resultString[0] = str1;
        return resultString;
    }

    else{
        for(int i = 0; i < n; i++){
            resultString[i]= str1.substring(0, 1) + interStr(str1.substring(1), str2.substring(1));

        }
    }
    return resultString;
}

    public static void main(String[] args) {
    Interleave obj = new Interleave();
    obj.interStr("12", "abc");
    for(int i = 0; i < obj.resultString.length; i ++){
        System.out.println(obj.resultString[i]);
    }

}

}

Answer 1

The question simply asked whether a recursive algorithm exists for the problem, and the answer is yes. To find it, look for the base case and then for the “step”.

The base case is when one of the two strings are empty:

• interleave(s1, "") = {s1}

• interleave("", s2) = {s2}

Notice the order of the arguments doesn’t really matter, because

• interleave("ab", "12") = {“ab12”, “a1b2”, “1ab2”, “a12b”, “1a2b”, “12ab”} = interleave("12", "ab")

So since the order doesn’t matter we’ll look at recursing on the length of the first string.

Okay so let’s see how one case leads to the next. I’ll just use a concrete example, and you can generalize this to real code.

• interleave("", "abc") = {“abc”}
• interleave("1", "abc") = {“1abc”, “a1bc”, “ab1c”, “abc1”}
• interleave("12", "abc") = {“12abc”, “1a2bc”, “1ab2c”, “1abc2”, “a12bc”, “a1b2c”, “a1bc2”, “ab12c”, “ab1c2” “abc12”}

So everytime we added a character to the first string, we formed the new result set by adding the new character to all possible positions in the old result set. Let’s look at exactly how we formed the third result above from the second. How did each element in the second result turn into elements in the third result when we added the “2”?

• “1abc” => “12abc”, “1a2bc”, “1ab2c”, “1abc2”
• “a1bc” => “a12bc”, “a1b2c”, “a1bc2”
• “ab1c” => “ab12c”, “ab1c2”
• “abc1” => “abc12”

Now look at things this way:

• “1abc” => {1 w | w = interleave(“2”, “abc”)}
• “a1bc” => {a1 w | w = interleave(“2”, “bc”)}
• “ab1c” => {ab1 w | w = interleave(“2”, “c”)}
• “abc1” => {abc1 w | w = interleave(“2”, “”)}

Although one or two examples doesn’t prove a rule in general, in this case you should be able to infer what the rule is. You will have a loop, with recursive calls inside it.

This is actually a little more fun to do with pure functional programming, but you tagged the question with Java.

Hopefully this is a start for you. If you get stuck further you can do a web search for “interleaving strings” or “interleaving lists”. There are some solutions out there.

EDIT:

Okay I just wrote the silly thing! It’s a lot of fun to write these things in scripting languages, so I thought it would be great to see what it looked like in Java. Not as bad as I thought it would be! Here it is, packaged as an entire Java application.

import java.util.ArrayList;
import java.util.List;

public class Interleaver {

    /**
     * Returns a list containing all possible interleavings of two strings.
     * The order of the characters within the strings is preserved.
     */
    public static List<String> interleave(String s, String t) {
        List<String> result = new ArrayList<String>();
        if (t.isEmpty()) {
            result.add(s);
        } else if (s.isEmpty()) {
            result.add(t);
        } else {
            for (int i = 0; i <= s.length(); i++) {
                char c = t.charAt(0);
                String left = s.substring(0, i);
                String right = s.substring(i);
                for (String u : interleave(right, t.substring(1))) {
                    result.add(left + c + u);
                }
            }
        }
        return result;
    }

    /**
     * Prints some example interleavings to stdout.
     */
    public static void main(String[] args) {
        System.out.println(interleave("", ""));
        System.out.println(interleave("a", ""));
        System.out.println(interleave("", "1"));
        System.out.println(interleave("a", "1"));
        System.out.println(interleave("ab", "1"));
        System.out.println(interleave("ab", "12"));
        System.out.println(interleave("abc", "12"));
        System.out.println(interleave("ab", "1234"));
    }
}