Question

New CS student, studying for a final. I am trying to figure out how many times a recursive method will be called in general. Added the code as an example. If I input abcd and efgh, how many calls based on the size of the Strings? If n is any data size, the # of calls is n(?) in any recursive method.

public static String interweave(String s1, String s2)
{ 
   if (s1.equals("") ) return s2;
   else if (s2.equals("")) return s1;
   else return "" + interweave(s1.substring(0,s1.length()-1), s2.substring(0,s2.length()-1))
       +s1.charAt(s1.length()-1)+s2.charAt(s2.length()-1);
}

Answer 1

Notice that in your question, at each recursive step you reduce by one the size of both strings until either one of them has length zero (the base cases of the recursion). It’s easy to see that the number of recursive calls is min(m, n) + 1, where m is the initial length of s1 and n is the initial length of s2.

For example, if s1 = "abc" and s2 = "de" it will take 2 recursive calls to traverse s2 (the string with the minimum length) plus one extra call for exiting at the base case, therefore min(s1.length(), s2.length()) + 1 == 3. You can test it programmatically like this:

static int counter;

public static String interweave(String s1, String s2) {
    counter++;
    if (s1.equals(""))
        return s2;
    else if (s2.equals(""))
        return s1;
    else
        return interweave(s1.substring(0, s1.length()-1), s2.substring(0, s2.length()-1))
        + s1.charAt(s1.length()-1)+s2.charAt(s2.length()-1);
}

public static int count(String s1, String s2) {
    counter = 0;
    interweave(s1, s2);
    return counter;
}

Now when you run the following statements, the formula works as expected:

// s1.length() == s2.length()
System.out.println(count("abcde", "fghij"));
> 6

// s1.length() > s2.length()
System.out.println(count("abcde", "fg"));
> 3

// s1.length() < s2.length()
System.out.println(count("ab", "cdefg"));
> 3