Develop Reference

I wanted to sort my duplicated characters in alphabetical order, what can I do?

import java.util.Scanner;
public class Chupapi {
static Scanner input = new Scanner(System.in);
public static void main(String[] args) {
String str = "";
System.out.print("Enter string: ");
str = input.nextLine();
getConsonantCount(str);
}
public static void getConsonantCount(String str) {
char string[] = str.toCharArray();
System.out.print("Duplicate characters in a given string: ");
for(int i = 0, count = 0; i <string.length; i++) {
count = 1;
for(int j = i+1; j <string.length; j++) {
if(string[i] == string[j] && string[i] != ' ') {
count++;
string[j] = '0';
}
}
if(count > 1 && string[i] != '0')
System.out.print(string[i]);
}
}
}
I wanted to sort out the "Duplicate characters in a given string: " output like in an alphabetical order example is,
"Duplicate characters in a given string: heuro"
I want it to be
"Duplicate characters in a given string: ehoru"
That's preatty much all that I want to do with this code, any tips on what should I do or use?

Use Arrays.sort() to sort the array before finding duplicate characters.
public static void getConsonantCount(String str)
{
char string[] = str.toCharArray();
Arrays.sort(string);
System.out.print("Duplicate characters in a given string: ");
for (int i = 0, count = 0; i < string.length; i++)
{
count = 1;
for (int j = i + 1; j < string.length; j++)
{
if (string[i] == string[j] && string[i] != ' ')
{
count++;
string[j] = '0';
}
}
if (count > 1 && string[i] != '0')
System.out.print(string[i]);
}
}

how to split a string by using charAt and string.length()

only allow charAt method and length method . Thank you so much!
void runApp() {
String str = "345, 688"; //->"345" "688"
String value = strCut(str);
}
String strCut(String str) {
int result = 0;
for (int i = 0; i < str.length(); i++) {
if (str.charAt(3) == ',') {
what should i write here ? ?
}

Your code needs some refactoring, try this:
void runApp() {
String str = "345, 688"; //->"345" "688"
String value = strCut(str);
}
String strCut(String str) {
int result = 0;
for (int i = 0; i < str.length(); i++) {
int cutStringIndex;
if (str.charAt(i) == ',') {
cutStringIndex = i;
}
for (int i = 0; i < cutStringIndex(); i++) {
String cutStringOne = "";
cutStringOne = cutStringOne + str.charAt(i);
}
for (int i = cutStringIndex() + 1; i < str.length(); i++) {
String cutStringTwo = "";
cutStringTwo = cutStringTwo + str.charAt(i);
}
cutString = cutStringOne + " " + cutStringTwo;
return cutString;
}
This will take out the comma which appears to be what you were looking for. I only used the two methods you asked for. Essentially this code gets the index of the comma, then reconstructs the two parts of the strings until it reaches the point of the comma, and skips over it. It may need some minor tweaks for your situation but this should be what you're looking for.

It can be done like this, Suppose String s="200,300,450,600" and you have to split given string using charAt() and string.length() method then first add ',' at the end of the string as given in the code below.
String s="200,300,450,600,",str="";
for(int i=0;i<s.length();i++){
char ch=s.charAt(i);
if(ch!=','){ //checking if particular character is not ','
str+=ch; //storing it in str string
}
else{
System.out.println(str); //printing each string before ',' is found
str="";
}
}
The output of above code will be:200
300
450
600(all the numbers will be printed on next line)

If you want to use only charAt and string.length() then you should try this
void runApp{
String str = "345, 688, 123";
String values[] = strCut(str); //values[0] = 345, values[1] = 688, values[2] = 123
for(String value : values){
System.out.print(value + " ");
}
}
String[] strCut(String str) {
int elements = 1;
int index = 0;
for(int i = 0; i < str.length(); i++){
if(str.charAt(i) == ',')
elements++;
}
String result[] = new String[elements];
for(int i = 0; i < result.length; i++)
result[i] = "";
for (int i = 0; i < str.length(); i++) {
if (str.charAt(i) != ',') {
if(str.charAt(i) != ' ')
result[index] = result[index] + str.charAt(i);
}
else index++;
}
return result;
}

You can do it as follows:
public class Main {
public static void main(String[] args) {
// Test
runApp();
}
static void runApp() {
String str = "345, 688"; // Expected->"345" "688"
String value = strCut(str);
System.out.println(value);// Display the result
}
static String strCut(String str) {
// Initialise result with a "
String result = "\"";
for (int i = 0; i < str.length(); i++) {
if (str.charAt(i) == ',') {// Check char at the index, i
// Add " at the end of one number and again " at the start of the next
result += "\" \"";
} else if (str.charAt(i) != ' ') {
result += str.charAt(i);
}
}
// Add " at the end
result += "\"";
// Finally, return result
return result;
}
}
Output:
"345" "688"

if you must want to make use of charAt() then do like below..
ArrayList<String> stringArr= new ArrayList<String>();
int startindex=0;
for (int i = 0; i < str.length(); i++)
{
if (str.charAt(i) == ',')
{
String partString = str.substring(startindex, i) ;
startindex=i+1;
stringArr.add(partString);
}
}
String lastString = str.substring(startindex, str.length()) ;
stringArr.add(lastString);
OR
You can simply use split method like below
String[] parts = string.split(",");
String part1 = parts[0]; // 345
String part2 = parts[1]; // 688

You can achieve it by simply doing this,
This will give you the desired result.
String str = "345,688";
ArrayList<String> stringArray = new ArrayList<>();
int startindex=0;
for (int i = 0; i < str.length(); i++) {
if(str.charAt(i) == ',') {
String subStr = str.substring(startindex, i);
startindex = i+1;
stringArray.add(subStr);
}
}
stringArray.add(str.substring(startindex));

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

How can I compare these two strings and remove a common letter without using arrays?

This was the code I designed to solve this problem but it seems not to work at all.I used nested for loops to compare the letters of the first string and the second string since they are likely to have different lengths
import java.util.*;
public class Trim
{
public static String myTrim(String input, String list)
{
String r = "";
for (int i = 1; i < input.length();i++)
{
for (int k = 1; k < list.length();k++)
{
if (input.charAt(i) != list.charAt(i))
{
r += input.charAt(i);
}
}
}
return r;
}
}

I guess you should use the method String.indexOf.
So:
public static String myTrim(String input, String list)
{
StringBuilder result = new StringBuilder();
char c;
for (int i = 0; i < input.length(); i++)
{
c = input.charAt(i);
if (list.indexOf(c) < 0)
result.append(c);
}
return result.toString();
}

Try using a flag to determine whether to character gets repeated or not:
String r = "";
for (int i = input.length() - 1; 0 <= i; i --) {
if (-1 == list.indexOf(input.charAt(i))) {
r += input.charAt(i);
}
}
OR
String r = "";
boolean found;
for (int i = input.length() - 1, j = list.length() - 1; 0 <= i; i--) {
found = false;
for (int k = j; 0 <= k; k--) {
if (list.charAt(k) == input.charAt(i)) {
found = true;
break;
}
}
if (!found) {
r += input.charAt(i);
}
}

We have to filter out the characters from input which appears in list.
Now we have to check whether each character of the input appears in the list or not.
The k value will be less then list.length() if the character of input present in the list string.
After the loop we check the k value and append it to the new string.
public static String myTrim(String input, String list)
{
String r = "";
for (int i = 0; i < input.length();i++)
{
int k = 0;
for (; k < list.length();k++)
{
if (input.charAt(i) == list.charAt(k))
{
break;
}
}
if(k == list.length())
r += input.charAt(i);
}
return r;
}

A nice one-liner solution would be to use Guava Charmatcher:
CharMatcher.anyOf(list).removeFrom(input);

I have tried this code and it's working fine with both of your inputs
for (int i = 0; i < S1.length(); i++) {
for(int j=0;j< S2.length();j++) {
if(S1.charAt(i)==S2.charAt(j)) {
char Temp= S2.charAt(j);
String Temp2=String.valueOf(Temp);
S1=S1.replace(Temp2, "");
}
}
}
This is code
import java.util.Scanner;
public class StringRemoveChar {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
String S1, S2;
S1 = scanner.nextLine();
S2 = scanner.nextLine();
for (int i = 0; i < S1.length(); i++) {
for (int j = 0; j < S2.length(); j++) {
if (S1.charAt(i) == S2.charAt(j)) {
char Temp = S2.charAt(j);
String Temp2 = String.valueOf(Temp);
S1 = S1.replace(Temp2, "");
System.out.println(S1.length());
}
}
}
System.out.println(S1);
}
}
Input:
Miyazaki
you
Output:
Miazaki

We can use replaceAll and use one loop over ,this will make the solution simple
public static String myTrim(String input, String list)
{
for(int i=0;i<list.length();i++)
{
input=input.replaceAll(list.charAt(i)+"","");
}
return input;
}
Input: myTrim("Miyazaki","you")
Output: Miazaki
Full code for reference
package stackoverflow.string;
public class StringManipulation
{
public static void main(String[] args)
{
System.out.println(myTrim("University of Miami","city"));
}
public static String myTrim(String input, String list)
{
for(int i=0;i<list.length();i++)
{
input=input.replaceAll(list.charAt(i)+"","");
}
return input;
}
}

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;
}
}