How to Compress a String using Recursion? (RLE algorithm)

Question

I'm having slight trouble trying to compress a string using recursion.

For example, consider the following string:

qwwwwwwwwweeeeerrtyyyyyqqqqwEErTTT

After applying the RLE algorithm, this string is converted into:

q9w5e2rt5y4qw2Er3T

In the compressed string, "9w" represents a sequence of 9 consecutive lowercase "w" characters. "5e" represents 5 consecutive lowercase "e" characters, etc.

I already have a code for compressing it without recursion:

import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Compress {

    public static String input(String source) {
        StringBuffer coding = new StringBuffer();
        for (int i = 0; i < source.length(); i++) {
            int runLength = 1;
            while (i+1 < source.length() && source.charAt(i) == source.charAt(i+1)) {

                runLength++;   

                i++;

           }
            if (runLength>1){
            coding.append(runLength);
            }
            coding.append(source.charAt(i));
        }
        return coding.toString();
    }

    public static void main(String[] args) {

        IO.outputStringAnswer("Enter a string");
        String str = IO.readString();
        String result = ""; 
        result=input(str); //function(variable)


        IO.outputStringAnswer(result);
    }
}

But I am unsure if this is able to be turned into a recursion version of this.

You certainly *can* make a recursive version of it, but there's no clear benefit to doing so. — Jim Mischel, Nov 25 '13 at 19:16
Oh, so I would have to rewrite the whole code? I'm still new to all this so I'm not very used to writing codes with recursion yet, sadly. — AChoi, Nov 25 '13 at 19:19
Why do you want to compress a string using recursion? I am having a hard time seeing a reason for doing so - at least in Java. — Vegard, Nov 25 '13 at 19:27
Oh it's part of an assignment I need to complete but I've never really used recursion in an assignment before until now. I've looked up some examples but I can't seem to figure out how to get recursion to be used for compression. — AChoi, Nov 25 '13 at 19:29
Does the assignment provide any guidance or justification for using recursion? This seems like a really bad use of recursion. — Jim Garrison, Nov 25 '13 at 19:34
Oh wow, I completely read the assignment wrong haha. They provided me a method file so I just need to complete that. I think I have the general idea of what to do now, thank you for the comments though ^^ — AChoi, Nov 25 '13 at 19:44

lex82 · Answer 1 · 2013-11-25T21:09:57.853

This is most likely what you are looking for:

public static String compress(String source) {
    if (source.length() <= 1) return source;

    int runLength = 1;
    while (runLength < source.length() && source.charAt(0) == source.charAt(runLength)) {
        runLength++;
    }

    String lengthString = runLength > 1 ? String.valueOf(runLength) : "";
    return lengthString + source.substring(0,1) + compress(source.substring(runLength));
}

I assume your source String does not contain any digits. As you can see the function calls itself recursively in the last line with the remainder of the source string.

Henry Florence · Answer 2 · 2013-11-28T00:06:06.270

This is an interesting question. One solution that uses no iteration at all, only recursion would be:

public static String compressRecursion(String str, char curChar, int curCount) {
    // termination case - reached end of the source string
    if(str.length() == 0)
        return "" + (curCount == 1 ? "" : curCount) + curChar;

    // branch on change in next character
    String nextStr = str.substring(1,str.length());
    if(str.charAt(0) == curChar)
        return compressRecursion(nextStr,curChar,curCount + 1);
    else
        return "" + (curCount == 1 ? "" : curCount) + curChar
                + compressRecursion(nextStr,str.charAt(0),1);
}
public static String compress(String source) {
    return compressRecursion(source, source.charAt(0), 0);
}

This probably wouldn't be much use in production code, as a 'Stack Overflow' exception will occur with an input of any reasonable length, this is because a new stack frame will be created for every function call and thus character in the input. Java is not really designed to run code like this.

A synonymous compress function written in scheme (which has no iteration constructs):

(define (num2str num)
  (if (= 1 num) "" (number->string num)))

(define (first-char str)
  (substring str 0 1))

(define (next-string str)
  (substring str 1 (string-length str)))

(define (compress str char count)
  (cond [(= 0 (string-length str)) (string-append (num2str count) char)]
        [(string=? char (first-char str))
          (compress (next-string str) char (+ count 1))]
        [ else 
          (string-append (num2str count) char
            (compress (next-string str) (first-char str) 1))]))

(define (compressStart str)
  (compress str (first-char str) 0))

A functional language like scheme will use tail recursion optimisation to prevent stack overflow,and makes a function call a much more lightweight operation than in an imperative language like Java.

How to Compress a String using Recursion? (RLE algorithm)

2 Answers2