String compression algorithm in Java

Question

I am looking to implement a method to perform basic string compression in the form of:

aabcccccaaa -> a2b1c5a3

I have this program:

import java.util.Scanner;

public class Main {

  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);

    String str = sc.nextLine();

    System.out.println(compress(str));
  }

  public static String compress(String str) {
    char[] chars = str.toCharArray();

    int count = 0;
    String result = "";

    for (int i = 0; i < chars.length; i++) {
      char curr = chars[i];
      result += curr;

      for (int j = i; j < chars.length; j++) {
        if (chars[j] == curr) {
          count++;
        }
        else {
          i += count;
          break;
        }
      }
      result += count;
      count = 0;
    }

    return result;
  }

}

But in my tests I am always missing the last character count.

I assume this is because the program gets out of the inner for loop before it should, but why is this the case?

Thanks a lot

yeah you need to keep track of the last character you end up coming across and when you exit your for loop add the a3 to your string — RAZ_Muh_Taz, Commented Dec 7, 2016 at 22:50
Running it with aaabbbccc gives a3b2c2c1. You need to rethink about your algorithm. In particular, about the way you increment i. — JB Nizet, Commented Dec 7, 2016 at 22:50
Do not build potentially long Strings "adding" to the current, immutable value of a variable of type String: use StringBuilder.append(). Try to use just one iteration. — greybeard, Commented Dec 7, 2016 at 22:50
A great way to figure this kind of thing out is to use a debugger — Simon, Commented Dec 7, 2016 at 22:52
Sounds like an "off by one error". You should use a debugger to help you track down the exact problem. Debugging skills are very important for any programmer. In fact, they are more important than knowing how to write code. — Code-Apprentice, Commented Dec 7, 2016 at 23:03

RAZ_Muh_Taz · Accepted Answer · 2016-12-07 23:00:26Z

You don't need two for loops for this and can do it in one go like so

    String str = "aaabbbbccccca";
    char[] chars = str.toCharArray();
    char currentChar = str.length() > 0 ? chars[0] : ' ';
    char prevChar = ' ';
    int count = 1;
    StringBuilder finalString = new StringBuilder();

    if(str.length() > 0)
    for(int i = 1; i < chars.length; i++)
    {
        if(currentChar == chars[i])
        {
            count++;
        }else{
            finalString.append(currentChar + "" + count);
            prevChar = currentChar;
            currentChar = chars[i];
            count = 1;
        }
    }

    if(str.length() > 0 && prevChar != currentChar)
        finalString.append(currentChar + "" + count);

    System.out.println(finalString.toString());

Output is: a3b4c5a1 for aaabbbbccccca

The code looks carelessly formatted. Why don't you just let your IDE format the code for you? — Roland Illig, Commented Jun 13, 2020 at 6:31

henryoats · Accepted Answer · 2017-08-06 05:15:11Z

Keep a track of character that you are reading and compare it with next character of the string. If it is different, reset the count.

public static void stringCompression (String compression) {
        String finalCompressedString = "";
        char current = '1';
        int count = 0;

        compression = compression + '1';

        for (int i = 0; i < compression.length(); i++) {
            if (compression.charAt(i) == current) {
                count = count + 1;
            } else {
                if (current != '1') 
                    finalCompressedString = finalCompressedString + (current + Integer.toString(count));
                    count = 1;
                    current = compression.charAt(i);
            }
        }
        System.out.println(finalCompressedString);
    }

(Do not build potentially long Strings "adding" to a current, immutable value of type String: use StringBuilder.append().) Does this work for "press1"? — greybeard, Commented Aug 6, 2017 at 5:55

Anubhav Mishra · Accepted Answer · 2020-06-12 14:31:50Z

My answer for String Compression in java. In this what i have done is and what you should have done is that , Keep a record of the characters that that are coming for a specific number of times, do so by comparing the current character with the next character , and when the current and the next character become unequal reset the value of count and repeat the whole process again for the next different character. Hope it helps!

import java.util.*;
public class Main {
    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        String str = sc.nextLine();
        int count = 0;
        for (int i=0; i<str.length(); ++i) {
            int j=i+1;
            count=1;
            while (j!=str.length() && str.charAt(i) == str.charAt(j)) {
                count += 1;
                j += 1;
                i += 1;
            }
            System.out.print(str.charAt(i));
            if (count > 1) {
                System.out.print(count);
            }
        }
    }
}

Thanks for improving your answer. I still don't like it though since your code is worse than the original code. The method name main does not appropriately describe what it is doing, and the count should be appended unconditionally. — Roland Illig, Commented Jun 13, 2020 at 6:28

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