Java compressing Strings

Question

I need to create a method that receives a String and also returns a String.

Ex input: AAABBBBCC

Ex output: 3A4B2C

Well, this is quite embarrassing and I couldn't manage to do it on the interview that I had today ( I was applying for a Junior position ), now, trying at home I made something that works statically, I mean, not using a loop which is kind of useless but I don't know if I'm not getting enough hours of sleep or something but I can't figure it out how my for loop should look like. This is the code:

public static String Comprimir(String texto){

    StringBuilder objString = new StringBuilder();

    int count;
    char match;

        count = texto.substring(texto.indexOf(texto.charAt(1)), texto.lastIndexOf(texto.charAt(1))).length()+1;
        match = texto.charAt(1);
        objString.append(count);
        objString.append(match);

    return objString.toString();
}

Thanks for your help, I'm trying to improve my logic skills.

Lammas

Does ABC get "compressed" to 1A1B1C or stay as ABC? What about AABC -> 2ABC? – Irmgard

ABC should return ABC. And AABC should return 2ABC. Thanks! – Lammas

In the input the same alphabet always come together or not. Means can input be of the format AAABBBCCCAACDD?? – Isoagglutination

The exam didn't specify. I'd like to do it the hard(best) way :). – Lammas

Can the input string contain numbers? – Cam

Answer 1

Loop through the string remembering what you last saw. Every time you see the same letter count. When you see a new letter put what you have counted onto the output and set the new letter as what you have last seen.

String input = "AAABBBBCC";

int count = 1;

char last = input.charAt(0);

StringBuilder output = new StringBuilder();

for(int i = 1; i < input.length(); i++){
    if(input.charAt(i) == last){
    count++;
    }else{
        if(count > 1){
            output.append(""+count+last);
        }else{
            output.append(last);
        }
    count = 1;
    last = input.charAt(i);
    }
}
if(count > 1){
    output.append(""+count+last);
}else{
    output.append(last);
}
System.out.println(output.toString());

Lamination

Answer 2

You can do that using the following steps:

• Create a HashMap
• For every character, Get the value from the hashmap -If the value is null, enter 1 -else, replace the value with (value+1)
• Iterate over the HashMap and keep concatenating (Value+Key)

Stagnant

I don't think this will work AAABBAAA will compress to 6A2B which you cannot uncompress!!! – Dowell

Make sure the HashMap is a LinkedHashMap though, as other implementations will mess up the key order. – Dkl

your solution won't work with string like that: "AAABBBCCAA" – Zela

Answer 3

• use StringBuilder (you did that)
• define two variables - previousChar and counter
• loop from 0 to str.length() - 1
• each time get str.charat(i) and compare it to what's stored in the previousChar variable
• if the previous char is the same, increment a counter
• if the previous char is not the same, and counter is 1, increment counter
• if the previous char is not the same, and counter is >1, append counter + currentChar, reset counter
• after the comparison, assign the current char previousChar
• cover corner cases like "first char"

Something like that.

Carping

Answer 4

The easiest approach:- Time Complexity - O(n)

public static void main(String[] args) {
    String str = "AAABBBBCC";       //input String
    int length = str.length();      //length of a String

    //Created an object of a StringBuilder class        
    StringBuilder sb = new StringBuilder(); 

    int count=1;   //counter for counting number of occurances

    for(int i=0; i<length; i++){
        //if i reaches at the end then append all and break the loop
        if(i==length-1){         
            sb.append(str.charAt(i)+""+count);
            break;
        }

        //if two successive chars are equal then increase the counter
        if(str.charAt(i)==str.charAt(i+1)){   
            count++;
        }
        else{
        //else append character with its count                            
            sb.append(str.charAt(i)+""+count);
            count=1;     //reseting the counter to 1
        }
   }

    //String representation of a StringBuilder object
    System.out.println(sb.toString());   

}

Palp

very nice & simple answer. – Fusty

more easier would be when we initialize count as 0, then we need only one if condition - for(int i=0;i<str.length();i++) { count++; if(i==str.length()-1 || str.charAt(i)!=str.charAt(i+1)) { compressedStr.append(str.charAt(i)).append(count); count=0; } } – Cestar

Answer 5

In the count=... line, lastIndexOf will not care about consecutive values, and will just give the last occurence.

For instance, in the string "ABBA", the substring would be the whole string.

Also, taking the length of the substring is equivalent to subtracting the two indexes.

I really think that you need a loop. Here is an example :

public static String compress(String text) {
    String result = "";

    int index = 0;

    while (index < text.length()) {
        char c = text.charAt(index);
        int count = count(text, index);
        if (count == 1)
            result += "" + c;
        else
            result += "" + count + c;
        index += count;
    }

    return result;
}

public static int count(String text, int index) {
    char c = text.charAt(index);
    int i = 1;
    while (index + i < text.length() && text.charAt(index + i) == c)
        i++;
    return i;
}

public static void main(String[] args) {
    String test = "AAABBCCC";
    System.out.println(compress(test));
}

Malan

Very nice, I'm going to study all the variables you guys are giving me. I don't want to go through that embarrassment never again. – Lammas

Answer 6

Please try this one. This may help to print the count of characters which we pass on string format through console.

import java.util.*;

public class CountCharacterArray {
   private static Scanner inp;

public static void main(String args[]) {
   inp = new Scanner(System.in);
  String  str=inp.nextLine();
   List<Character> arrlist = new ArrayList<Character>();
   for(int i=0; i<str.length();i++){
       arrlist.add(str.charAt(i));
   }
   for(int i=0; i<str.length();i++){
       int freq = Collections.frequency(arrlist, str.charAt(i));
       System.out.println("Frequency of "+ str.charAt(i)+ "  is:   "+freq); 
   }
     }    
}

Bertrand

Answer 7

This is just one more way of doing it.

public static String compressor(String raw) {
        StringBuilder builder = new StringBuilder();
        int counter = 0;
        int length = raw.length();
        int j = 0;
        while (counter < length) {
            j = 0;
            while (counter + j < length && raw.charAt(counter + j) == raw.charAt(counter)) {
                j++;
            }

            if (j > 1) {
                builder.append(j);
            }
            builder.append(raw.charAt(counter));
            counter += j;
        }

        return builder.toString();
    }

Wychelm

Thanks to you mate, I hope you have a nice day. You just got out of bed and you already helped a guy :) – Lammas

Answer 8

Java's not my main language, hardly ever use it, but I wanted to give it a shot :] Not even sure if your assignment requires a loop, but here's a regexp approach:

 public static String compress_string(String inp) {
      String compressed = "";
      Pattern pattern = Pattern.compile("([\\w])\\1*");
      Matcher matcher = pattern.matcher(inp);
      while(matcher.find()) {
         String group = matcher.group();
         if (group.length() > 1) compressed += group.length() + "";
         compressed += group.charAt(0);
      }
      return compressed;
   }

Irmgard

Answer 9

The following can be used if you are looking for a basic solution. Iterate through the string with one element and after finding all the element occurrences, remove that character. So that it will not interfere in the next search.

public static void main(String[] args) {
    String string = "aaabbbbbaccc";
    int counter;
    String result="";
    int i=0;
    while (i<string.length()){
        counter=1;
        for (int j=i+1;j<string.length();j++){ 
            System.out.println("string length ="+string.length());  
            if (string.charAt(i) == string.charAt(j)){
                  counter++;
            }
      }
      result = result+string.charAt(i)+counter; 
      string = string.replaceAll(String.valueOf(string.charAt(i)), ""); 
    }
    System.out.println("result is = "+result);
}

And the output will be := result is = a4b5c3

Pixie

Answer 10

private String Comprimir(String input){
        String output="";
        Map<Character,Integer> map=new HashMap<Character,Integer>();
        for(int i=0;i<input.length();i++){
            Character character=input.charAt(i);
            if(map.containsKey(character)){
                map.put(character, map.get(character)+1);
            }else
                map.put(character, 1);
        }
        for (Entry<Character, Integer> entry : map.entrySet()) {
            output+=entry.getValue()+""+entry.getKey().charValue();
        }
        return output;
    }

One other simple way using Multiset of guava-

import java.util.Arrays;

import com.google.common.collect.HashMultiset;
import com.google.common.collect.Multiset;
import com.google.common.collect.Multiset.Entry;

public class WordSpit {
    public static void main(String[] args) {
        String output="";
        Multiset<String> wordsMultiset = HashMultiset.create();
        String[] words="AAABBBBCC".split("");
        wordsMultiset.addAll(Arrays.asList(words));
        for (Entry<String> string : wordsMultiset.entrySet()) {
            if(!string.getElement().isEmpty())
                output+=string.getCount()+""+string.getElement();
        }
        System.out.println(output);
    }
}

Beady

Great way of solving it, I'll analyze it step by step. Thanks for your time! – Lammas

It doesn't work because it won't handle character repeats: "AAABBBCCAA" would be "5A3B2C" instead of "3A3B2C2A". Moreover, HashMap does not preserve key order. Use LinkedHashMap instead. – Dkl

Answer 11

consider the below Solution in which the String s1 identifies the unique characters that are available in a given String s (for loop 1), in the second for loop build a string s2 that contains unique character and no of times it is repeated by comparing string s1 with s.

public static void main(String[] args) 
{
    // TODO Auto-generated method stub

    String s = "aaaabbccccdddeee";//given string
    String s1 = ""; // string to identify how many unique letters are available in a string
    String s2=""; //decompressed string will be appended to this string
    int count=0;
    for(int i=0;i<s.length();i++) {
        if(s1.indexOf(s.charAt(i))<0) {
            s1 = s1+s.charAt(i);
        }
    }
    for(int i=0;i<s1.length();i++) {
        for(int j=0;j<s.length();j++) {
            if(s1.charAt(i)==s.charAt(j)) {
                count++;
            }
        }
        s2=s2+s1.charAt(i)+count;
        count=0;
    }

    System.out.println(s2);
}

Interminable

:-P added explanation – Interminable

Answer 12

It may help you.

public class StringCompresser
{
public static void main(String[] args)
{
    System.out.println(compress("AAABBBBCC"));
    System.out.println(compress("AAABC"));
    System.out.println(compress("A"));
    System.out.println(compress("ABBDCC"));
    System.out.println(compress("AZXYC"));
}

static String compress(String str)
{
    StringBuilder stringBuilder = new StringBuilder();
    char[] charArray = str.toCharArray();
    int count = 1;
    char lastChar = 0;
    char nextChar = 0;
    lastChar = charArray[0];
    for (int i = 1; i < charArray.length; i++)
    {
        nextChar = charArray[i];
        if (lastChar == nextChar)
        {
            count++;
        }
        else
        {
            stringBuilder.append(count).append(lastChar);
            count = 1;
            lastChar = nextChar;

        }
    }
    stringBuilder.append(count).append(lastChar);
    String compressed = stringBuilder.toString();

    return compressed;
} 
}

Output:

3A4B2C
3A1B1C
1A
1A2B1D2C
1A1Z1X1Y1C

Bilingual

Answer 13

The answers which used Map will not work for cases like aabbbccddabc as in that case the output should be a2b3c2d2a1b1c1.

In that case this implementation can be used :

private String compressString(String input) {
        String output = "";
        char[] arr = input.toCharArray();
        Map<Character, Integer> myMap = new LinkedHashMap<>();
        for (int i = 0; i < arr.length; i++) {
            if (i > 0 && arr[i] != arr[i - 1]) {
                output = output + arr[i - 1] + myMap.get(arr[i - 1]);
                myMap.put(arr[i - 1], 0);
            }
            if (myMap.containsKey(arr[i])) {
                myMap.put(arr[i], myMap.get(arr[i]) + 1);
            } else {
                myMap.put(arr[i], 1);
            }
        }

        for (Character c : myMap.keySet()) {
            if (myMap.get(c) != 0) {
                output = output + c + myMap.get(c);
            }
        }

        return output;
    }

Tempest

Answer 14

O(n) approach

No need for hashing. The idea is to find the first Non-matching character. The count of each character would be the difference in the indices of both characters.

for a detailed answer: https://mcmap.net/q/821821/-in-place-run-length-encoding-algorithm

The only catch is that we need to add a dummy letter so that the comparison for the last character is possible.

private static String compress(String s){
    StringBuilder result = new StringBuilder();
    int j = 0;
    s = s + '#';
    for(int i=1; i < s.length(); i++){
        if(s.charAt(i) != s.charAt(j)){
            result.append(i-j);
            result.append(s.charAt(j));
            j = i;
        }
    }
   return result.toString();
}

Dovetail

for taking care of last characters, you can just add a special case before returning the string, look out for my answer. Although your answer is correct, you really don't need # and all, just have simple counter and reset it, whenever the current and next character changes. – Defamation

Answer 15

 // O(N) loop through entire character array
 // match current char with next one, if they matches count++
 // if don't then just append current char and counter value and then reset counter.
// special case is the last characters, for that just check if count value is > 0, if it's then append the counter value and the last char

 private String compress(String str) {
        char[] c = str.toCharArray();
        String newStr = "";
        int count = 1;
        for (int i = 0; i < c.length - 1; i++) {
            int j = i + 1;
            if (c[i] == c[j]) {
                count++;
            } else {
                newStr = newStr + c[i] + count;
                count = 1;
            }
        }

        // this is for the last strings...
        if (count > 0) {
            newStr = newStr + c[c.length - 1] + count;
        }

        return newStr;
    }

Defamation

@Cristian, although my answer prints, A3C4, but you can change it easily by changing newStr. It's also solving this problem in O(N) time. – Defamation

Answer 16

This is a leet code problem 443. Most of the answers here uses StringBuilder or a HashMap, the actual problem statement is to solve using the input char array and in place array modification.

public int compress(char[] chars) {
    int startIndex = 0;
    int lastArrayIndex = 0;
    if (chars.length == 1) {
      return 1;
    }
    if (chars.length == 0) {
      return 0;
    }
    for (int j = startIndex + 1; j < chars.length; j++) {
      if (chars[startIndex] != chars[j]) {

        chars[lastArrayIndex] = chars[startIndex];
        lastArrayIndex++;
        if ((j - startIndex) > 1) {
          for (char c : String.valueOf(j - startIndex).toCharArray()) {
            chars[lastArrayIndex] = c;
            lastArrayIndex++;
          }
        }
        startIndex = j;
      }
      if (j == chars.length - 1) {
        if (j - startIndex >= 1) {
          j = chars.length;
          chars[lastArrayIndex] = chars[startIndex];
          lastArrayIndex++;
          for (char c : String.valueOf(j - startIndex).toCharArray()) {
            chars[lastArrayIndex] = c;
            lastArrayIndex++;
          }
        } else {
          chars[lastArrayIndex] = chars[startIndex];
          lastArrayIndex++;
        }
      }
    }
    return lastArrayIndex;
  }
}

Phylogeny

Answer 17

The code below will ask the user for user to input a specific character to count the occurrence .

import java.util.Scanner;

class CountingOccurences {

public static void main(String[] args) {

    Scanner inp = new Scanner(System.in);

    String str;
    char ch;
    int count=0;

    System.out.println("Enter the string:");
    str=inp.nextLine();
    System.out.println("Enter th Char to see the occurence\n");
    ch=inp.next().charAt(0);

    for(int i=0;i<str.length();i++)
    {
                if(str.charAt(i)==ch)
        {
            count++;
                }
    }

        System.out.println("The Character is Occuring");
        System.out.println(count+"Times");


}

}

Vendee

Answer 18

public static char[] compressionTester( char[] s){

    if(s == null){
        throw new IllegalArgumentException();
    }

    HashMap<Character, Integer> map = new HashMap<>();
    for (int i = 0 ; i < s.length ; i++) {

        if(!map.containsKey(s[i])){
            map.put(s[i], 1);
        }
        else{
            int value = map.get(s[i]);
            value++;
            map.put(s[i],value);
        }           
    }               
    String newer="";

    for( Character n : map.keySet()){

        newer = newer + n + map.get(n); 
    }
    char[] n = newer.toCharArray();

    if(s.length > n.length){
        return n;
    }
    else{

        return s;               
    }                       
}

Aerology

Answer 19

package com.tell.datetime;

import java.util.Stack;
public class StringCompression {
    public static void main(String[] args) {
        String input = "abbcccdddd";
        System.out.println(compressString(input));
    }

    public static String compressString(String input) {

        if (input == null || input.length() == 0)
            return input;
        String finalCompressedString = "";
        String lastElement="";
        char[] charArray = input.toCharArray();
        Stack stack = new Stack();
        int elementCount = 0;
        for (int i = 0; i < charArray.length; i++) {
            char currentElement = charArray[i];
            if (i == 0) {
                stack.push((currentElement+""));
                continue;
            } else {
                if ((currentElement+"").equalsIgnoreCase((String)stack.peek())) {
                    stack.push(currentElement + "");
                    if(i==charArray.length-1)
                    {
                        while (!stack.isEmpty()) {

                            lastElement = (String)stack.pop();
                            elementCount++;
                        }

                        finalCompressedString += lastElement + "" + elementCount;
                    }else
                    continue;
                }

                else {
                    while (!stack.isEmpty()) {

                        lastElement = (String)stack.pop();
                        elementCount++;
                    }

                    finalCompressedString += lastElement + "" + elementCount;
                    elementCount=0;
                    stack.push(currentElement+"");
                }

            }
        }

        if (finalCompressedString.length() >= input.length())
            return input;
        else
            return finalCompressedString;
    }

}

Kinlaw

Answer 20

public class StringCompression {
    public static void main(String[] args){
        String s = "aabcccccaaazdaaa";

        char check = s.charAt(0);
        int count = 0;

        for(int i=0; i<s.length(); i++){
            if(s.charAt(i) == check) {
                count++;
                if(i==s.length()-1){
                System.out.print(s.charAt(i));
                System.out.print(count);
             }
            } else {
                System.out.print(s.charAt(i-1));
                System.out.print(count);
                check = s.charAt(i);
                count = 1;
                if(i==s.length()-1){
                    System.out.print(s.charAt(i));
                    System.out.print(count);
                 }
            }
        }
    }

Bela

Please don't just dump code as an answer, add a description on how this solves the question. – Alvertaalves

Mark Rotteveel gave me -1. But the answer is valid. It will print the correct answer. – Bela

Yes, I gave you a -1, because you shouldn't just dump code, you should explain your solution. – Alvertaalves

Answer 21

public class StringCompression {
    public static void main(String... args){
        String s="aabbcccaa";
        //a2b2c3a2

        for(int i=0;i<s.length()-1;i++){
            int count=1;
            while(i<s.length()-1 && s.charAt(i)==s.charAt(i+1)){
                count++;
                i++;
            }
            System.out.print(s.charAt(i));
            System.out.print(count);
        }
        System.out.println(" ");
    }
}

Luminiferous

this will output a1b1 for abc. – Catharine

Answer 22

String input = "AAABBAABBCC";
        char[] inputArr = input.toCharArray();
        Map<String,Integer> mapCounter = new HashMap<String,Integer>();
        String resultant = "";
        String c1 = null;
        for(Character c: inputArr) {
            
            if(mapCounter.isEmpty()) {
                mapCounter.put(c.toString(), 1);
                continue;
            }
            if(mapCounter.keySet().stream().findFirst().isPresent()) {
                 c1 = mapCounter.keySet().stream().findFirst().get();
            }
            if(c1!=null && c1.equals(c.toString())) {
                if(mapCounter.containsKey(c.toString())) {
                    mapCounter.put(c.toString(), mapCounter.get(c.toString())+1);
                }
            }   
            else {
                resultant += c1.toString()+String.valueOf(mapCounter.get(c1));
                mapCounter.put(c.toString(), 1);
                mapCounter.remove(c1);
            }
            
        }
        resultant += mapCounter.keySet().stream().findFirst().get().toString()+String.valueOf(mapCounter.get(mapCounter.keySet().stream().findFirst().get()));
        
        System.out.println("Result::"+resultant);

Katushka