Develop Reference

Reverse encoded string

i am asking for help on how i would go about reversing my code so that the input 'A2B5C2' will give me the output 'AABBBBBCC', any suggestions?
Thanks
public static void printRLE(String str) {
int n = str.length();
for (int i = 0; i < n; i++) {
// Count occurrences of current character
int count = 1;
while (i < n - 1 && str.charAt(i) == str.charAt(i + 1)) {
count++;
i++;
}
// Print character and its count
System.out.print(str.charAt(i));
System.out.print(count);
}
}
public static void main(String[] args) {
String str = "AABBBBBCC";
printRLE(str);
}

To get the case, the number will more than 9, I'd suggest a simple regex to match letter+number, then just repeat the letter the number of times you need :
static String getRevRLE(String str) {
StringBuilder res = new StringBuilder();
Matcher m = Pattern.compile("([a-zA-Z][0-9]+)").matcher(str);
while (m.find()) {
String g = m.group();
res.append(g.substring(0, 1).repeat(Integer.parseInt(g.substring(1))));
}
return res.toString();
}
Using the Streams API you can reduce to
static String getRevRLE(String str) {
return Pattern.compile("([a-zA-Z][0-9]+)").matcher(str).results()
.map(MatchResult::group)
.map(g -> g.substring(0, 1).repeat(Integer.parseInt(g.substring(1))))
.collect(Collectors.joining());
}
Testing
public static void main(String[] args) {
String str = "AABBBBBCCCCCCCCCCCCCCCCCCCC";
String rle = getRLE(str);
String res = getRevRLE(rle);
System.out.println(res + " " + res.equals(str)); // AABBBBBCCCCCCCCCCCCCCCCCCCC true
}

Here you go:
public static String encode(String input) {
String output = "";
while (true) {
char c = input.charAt(0);
String countStr = "";
char current;
for (int i = 1; i < input.length() && Character.isDigit(current = input.charAt(i)); i++)
countStr += current;
int count = Integer.parseInt(countStr);
for (int i = 0; i < count; i++)
output += c;
int trimLength = 1 + countStr.length();
if (trimLength >= input.length())
return output;
else
input = input.substring(trimLength);
}
}

You can do this task like this:
public static String printRLE(String str) {
int n = str.length();
String result = "";
for (int i = 0; i < n - 1; i++) {
char ch = str.charAt(i);
if (!Character.isDigit(ch)) {
if (Character.isDigit(str.charAt(i + 1))) {
int fi = i + 1;
i += 2;
while (i < n && Character.isDigit(str.charAt(i))) i++;
int repeat = Integer.parseInt(str.substring(fi, i));
result += String.valueOf(ch).repeat(repeat);
i--;
} else result += ch;
}
}
return result;
}
public static void main(String[] args) {
String str = "10A10B32C1";
System.out.println(printRLE(str));
}
, output
AAAAAAAAAABBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBC

Hyphen between each integer in output

So I had to write a program that mimics a phone keypad, whereas it would convert a string of text to integers: abc(2), def(3), ghi(4), jkl(5), mno(6),
pqrs(7), tuv(8), wxyz(9). Except the output has to have hyphens(-) between the digits.
Example input: Alabama
Output: 2-5-2-2-2-6-2
But I just only output 2522262. How would I go about formatting this correctly?
public class Keypad {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.print("Enter a string: ");
String s = sc.nextLine();
System.out.println(getNumbers(s));
}
public static String getNumbers(String s) {
String result = new String();
for (int i = 0; i < s.length(); i++) {
if (Character.isLetter(s.charAt(i))) {
result += getNumber(Character.toUpperCase(s.charAt(i)));
}
else {
result += s.charAt(i);
}
}
return result;
}
public static int getNumber(char upperCaseLetter) {
int number = ((upperCaseLetter - 'A') / 3) + 2;
if (number < 7) {
return number;
}
else if (upperCaseLetter - 'A' < 20) {
return 7;
}
else if (upperCaseLetter - 'A' < 23) {
return 8;
}
else {
return 9;
}
}
}

Go to the place you construct the result and add the hyphen:
result += getNumber(Character.toUpperCase(s.charAt(i)));
result += "-";
Then before you return you will have to strip off the last hyphen:
return result.substring(0, result.length() - 1);
So the whole method would look like this:
public static String getNumbers(String s) {
String result = new String();
for (int i = 0; i < s.length(); i++) {
if (Character.isLetter(s.charAt(i))) {
result += getNumber(Character.toUpperCase(s.charAt(i)));
result += "-";
}
else {
result += s.charAt(i);
}
}
return result.substring(0, result.length() - 1);
}

StringJoiner was added to Java 8 for this purpose. Very simple and straightforward to use:
StringJoiner sj = new StringJoiner("-", "", "");
sj.add("1").add("1").add("2");
String desiredString = sj.toString();
or with Stream API, which might be a little more convenient in your case:
List<Integer> integers = Arrays.asList(1,2,3,4,5);
String hyphenSeparatedNumbers = integers.stream()
.map(Object::toString)
.collect(Collectors.joining("-"));
Also String.join is a superb alternative for this task.

There is a method in Java 8 that does just this. Use String.join, docs, to add a dash after each character.
public static String getNumbers(String s) {
String result = new String();
for (int i = 0; i < s.length(); i++) {
if (Character.isLetter(s.charAt(i))) {
result += getNumber(Character.toUpperCase(s.charAt(i)));
}
else {
result += s.charAt(i);
}
}
return String.join("-", result.split("");
}
Note
You should try to avoid using the += with strings, StringBuffer provides better performance. When you concatenate strings you are actually creating new objects for each new string you are concatenating. Imagine a large loop you will have n objects to create the new string.

Change the code in getNumbers(String) to
for (int i = 0; i < s.length(); i++) {
if (Character.isLetter(s.charAt(i))) {
result += getNumber(Character.toUpperCase(s.charAt(i)));
if (i < (s.length-1)
result += '-";
}
}
return result;
}

String compression algorithm in Java

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

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

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);
}

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);
}
}
}
}

java sort string array according to kurdish characters

Is there any short way to sort a string array by Kurdish characters? I've looked at some source on internet but I couldn't find any solution. There is a way to sort. Writing a code alike a novel but it is a very long work.
kurdish characters: a,b,c,ç,d,e,ê,f,g,h,i,î,j,k,l,m,n,o,p,q,r,s,ş,t,û,u,v,w,x,y,z

The Collator class should come in-handy here. To quote from the doc,
The Collator class performs locale-sensitive String comparison. You use this class to build searching and sorting routines for natural language text.
So try something like this:
Collator unicodeCollator = Collator.getInstance(Locale.UNICODE_LOCALE_EXTENSION);
Collections.sort(yourListOfCharacters, unicodeCollator);
Note that we are able to call java.util.Collections.sort directly as above, because Collator implements the Comparator interface.
If for whatever reasons Locale.UNICODE_LOCALE_EXTENSION doesn't work, here's the full list of supported locales. And you can create your own locale using the Locale constructor.

I've solved my problem: content of my file was like this:
*Nîzamettîn Ariç - Kardeş Türküler - Rojek Tê
Bê xem bê şer welat azad rojek tê
Rojek ronahî rojek bişahî rojek tê
Roj Roja me ye....
*Koma Çiya - Tolhildan ^ Daketine Meydanê
Daketine meydanê gerilayên dînemêr
Ji bona tolhildanê wek baz û piling û şêr...
My solution: thîs letters is proper for toLowerCase function:
ABCÇDEÊFGĞHİÎJKLMNOÖPQRSŞTÛUÜVWXYZ
just I was problem. because lowerCase(I) for turkish is ı; but for kurdish it is i.
code:
in onCreate():
...
alfabetBike();
...
public static void alfabetBike() {
for (int i = 0; i < tips.length(); i++) {
String[] derbasi_arr = sernavs[i];
String[] derbasi_got = gotins[i];
for (int j = 0; j < hejmar[i] - 1; j++) {
int indeks = j;
String yaMezin = derbasi_arr[j];
for (int k = j + 1; k < hejmar[i]; k++) {
if (compareTwoString(yaMezin.substring(1), derbasi_arr[k].substring(1)) > 1) {
yaMezin = derbasi_arr[k];
indeks = k;
}
}
if (indeks != j) {
derbasi_arr[indeks] = derbasi_arr[j];
String derbasi = derbasi_got[indeks];
derbasi_got[indeks] = derbasi_got[j];
derbasi_arr[j] = yaMezin;
derbasi_got[j] = derbasi;
}
}
gotins[i] = derbasi_got;
sernavs[i] = derbasi_arr;
}
}
private static void printFile(){
alfabetBike();
File root = android.os.Environment.getExternalStorageDirectory();
File dir = new File (root.getAbsolutePath() + "/alfabetfolder");
dir.mkdirs();
File file = new File(dir, "alfabet_title.txt");
File file2 = new File(dir, "alfabet.txt");
try {
FileOutputStream f = new FileOutputStream(file,false);
PrintWriter pw = new PrintWriter(f);
FileOutputStream f2 = new FileOutputStream(file2,false);
PrintWriter pw2 = new PrintWriter(f2);
for (int i = 0; i < tips.length(); i++) {
for (int j = 0; j < hejmar[i]; j++) {
Log.d("ssdddddd", "add" + hejmar[i] + "-" + j + " " + sernavs[i][j].trim());
pw.println(sernavs[i][j]);
pw.flush();
pw2.println(sernavs[i][j] + "\n" + gotins[i][j].trim());
pw2.flush();
}
}
pw.close();
f.close();
} catch (FileNotFoundException e) {
e.printStackTrace();
Log.i("erroooor", "******* File not found. Did you" +
" add a WRITE_EXTERNAL_STORAGE permission to the manifest?");
} catch (IOException e) {
e.printStackTrace();
}
}
public static int compareTwoString(String yek, String du) {
String d1 = yek, d2 = du;
d1 = strLower(d1, d1.charAt(0));
d2 = strLower(d2, d2.charAt(0));
int length, yaDirej;
if (yek.length() > du.length()) {
yaDirej = 1;
length = yek.length();
} else if (yek.length() < du.length()) {
yaDirej = 2;
length = du.length();
} else {
yaDirej = 0;
length = yek.length();
}
for (int i = 0; i < length; i++) {
int id1 = -1, id2 = -1;
if (i == d1.length() || i == du.length()) {
return yaDirej;
}
for (int j = 0; j < tips.length(); j++) {
if (d1.charAt(i) == tips.charAt(j)) id1 = j;
if (d2.charAt(i) == tips.charAt(j)) id2 = j;
}
if (id1 > id2)
return 2;
else if (id2 > id1)
return 1;
else
continue;
}
return 0;
}
public static String strLower(String str, char ziman){
final StringBuilder mutable = new StringBuilder(str);
final StringBuilder yedek = new StringBuilder(str.toLowerCase());
for (int i = 0; i < str.length(); i++) {
if (ziman == '?' && mutable.charAt(i) == 'I')
mutable.setCharAt(i, 'i');
else if (ziman == '*' && mutable.charAt(i) == 'I')
mutable.setCharAt(i, 'ı');
else mutable.setCharAt(i,yedek.charAt(i));
}
return mutable.toString();
}
edit:
in AndroidManifest.xml
<manifest...>
<uses-permission android:name="android.permission.WRITE_EXTERNAL_STORAGE" />
....
</manifest...>

You can build your own comparison so that, no matter what characters you are dealing with, it is going to sort the way you want. As you can see from the following code, I have set the comparison value by counting from a-z so that a=0, b=1...etc Then, I used the bubble sort strategy, which is basically switching the smallest elements continuously to the left and shifting others to the right.
public class Sort {
public static String compare(String compare1, String compare2) {
for (int i = 0; i < compare1.length(); i++) {
if (letterValue(compare1, i) < letterValue(compare2, i)) {
return compare1;
} else if (letterValue(compare1, i) > letterValue(compare2, i)) {
return compare2;
} else if (letterValue(compare1, i) == -1 || letterValue(compare2, i) == -1) {
System.out.print("Some letters are not within the alphabet!");
}
}
return compare1;
}
public static boolean smaller(String compare1, String compare2) {
if (compare(compare1, compare2).equalsIgnoreCase(compare1)) {
return true;
} else {
return false;
}
}
public static int letterValue(String input, int letterPosition) {
String order = "abcçdeêfghiîjklmnopqrsştûuvwxyz";
int value = -1;
for (int i = 0; i < order.length(); i++) {
if (input.toLowerCase().charAt(letterPosition) == order.charAt(i)) {
value = i;
}
}
return value;
}
public static void main(String[] args) {
String[] input = {"BARÊZ", "ÇÊneR", "ASTÛ", "badîn", "BADÎN"};
String swap;
int i, d;
for (i = 0; i < (input.length - 1); i++) {
for (d = 0; d < input.length - i - 1; d++) {
if (!smaller(input[d], input[d + 1])) {
swap = input[d];
input[d] = input[d + 1];
input[d + 1] = swap;
}
}
}
System.out.println("Sorted list: ");
for (i = 0; i < input.length; i++) {
System.out.print(input[i] + " ");
}
}
}
Output
Sorted list:
ASTÛ badîn BADÎN BARÊZ ÇÊneR

Java compressing Strings

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.

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());

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)

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.

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());
}

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));
}

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);
}
}
}

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;
}

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();
}

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

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);
}
}

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);
}

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

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;
}

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://stackoverflow.com/a/55898810/7972621
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();
}

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");
}
}

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;
}
}

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;
}
}

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);
}
}
}
}

// 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;
}

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(" ");
}
}

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;
}
}