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
I'm trying to write a method which takes a String, looks for Ints in it
and then adds them together.
for example:
String s = "five5six66"
should return 5+66 = 71 or:
String s = "1ciao2three3"
should return 1+2+3 = 6
The following is what I wrote already, but when I run it I get a
NumberFormatException
code (Update 1):
public static int addNumbers(String s) {
String numbers="";
int addNumbers = 0;
int i;
char c;
for (i=0; i<s.length(); i++) {
if (s.charAt(i)>='0' && s.charAt(i)<='9') {
c = s.charAt(i);
while (i<s.length()-1) {
if (s.charAt(i+1)>='0' && s.charAt(i+1)<='9')
numbers = numbers.valueOf(c + s.charAt(i+1));
addNumbers = addNumbers + Integer.parseInt(numbers);
}
addNumbers = addNumbers + Character.getNumericValue(c);
}
}
return addNumbers;
}
Hopefully you can help me fix this code and please, let me understand what I did wrong!
Also can I expand it so if I have a String like:
String s = "hi123and27"
I can get 123+27 = 150?
Because my code is limited to a 2 digit number as it is now.
I would suggest using REGEX to address your requirements:
you will need:
the REGEX pattern: "\d+"
an accumulator that is concatenating the value you get of the given String
Example:
public static void main(String[] args) {
String s = "hi123and27";
Pattern p = Pattern.compile("\\d+");
Matcher m = p.matcher(s);
int accumulator = 0;
while (m.find()) {
accumulator += Integer.parseInt(m.group());
}
System.out.println("final result: " + accumulator );
}
Regex + Java 8 streams:
public static int addNumbers(String str) {
return Arrays.stream(str.replaceAll("[^0-9]", " ").split(" "))
.filter(s -> !s.isEmpty())
.mapToInt(Integer::parseInt)
.sum();
}
EDIT regarding recommendations is the comments:
public static int addNumbers(String str) {
return Arrays.stream(str.split("[^0-9]+"))
.filter(s -> !s.isEmpty())
.mapToInt(Integer::parseInt)
.sum();
}
Try this
public static void main(String [] args){
String string = "hi123and27";
int size = string.length();
int sum = 0;
StringBuilder val = new StringBuilder();
for (int idx = 0; idx < size; idx++) {
Character character = string.charAt(idx);
if (Character.isDigit(character)) {
val.append(character);
//if last character is a digit
if((idx+1 == size) && val.length() > 0)
sum += Integer.parseInt(val.toString());
}else{
if(val.length() > 0)
sum += Integer.parseInt(val.toString());
//reset the val for digits between characters for it to store the next sequence
val.setLength(0);
}
}
System.out.println("The sum is : " + sum);
}
You should try this one.
public static int addNum(String text){
String numbers = "";
int finalResult = 0;
for(int i=0;i < text.length();i++){
if(isNumeric(text.substring(i, i + 1)))
{
numbers += text.substring(i, i + 1);
if(i==text.length()-1) {
finalResult += Integer.parseInt(numbers);
}
}else {
if(!numbers.equals("")){
finalResult += Integer.parseInt(numbers);
numbers = "";
}
}
}
return finalResult;
}
public static boolean isNumeric(String str)
{
try{
int d = Integer.parseInt(str);
}
catch(NumberFormatException ex){
return false;
}
return true;
}
I've been trying to use an array but it just seems to return the original String.
public static String capitalizeEveryOtherWord(String x) {
x = x.toLowerCase();
x.trim();
String[] words = x.split(" ");
for(int c = 2; c < words.length; c += 2)
words[c].toUpperCase();
return Arrays.toString(words);
}
Could anyone help?
toUpperCase() and trim() return new strings instead of modifying existing ones. You need to assign those new strings to something.
public static String capitalizeEveryOtherWord(String x) {
x = x.toLowerCase();
x = x.trim();
String[] words = x.split(" ");
for (int c = 2; c < words.length; c += 2)
words[c] = words[c].toUpperCase();
return Arrays.toString(words);
}
Also, you probably meant to start at index 0 or 1 – the first or second element, respectively.
Minitech has correctly identified the problem IMHO, but I would use a different regex-based approach:
public static String capitalizeEveryOtherWord(String x) {
StringBuilder result = new StringBuilder(x);
Matcher matcher = Pattern.compile("^ *\\w|\\w* \\w+ \\w").matcher(x);
while(matcher.find())
result.setCharAt(matcher.end() - 1, Character.toUpperCase(x.charAt(matcher.end() - 1)));
return result.toString();
}
(Tested and works).
This also works:
public class answerForStackOverflow {
public static void main(String[] args) {
String examplestring = "iwouldreallyliketothankforallthehelponstackoverflow";
String output = "";
for (int i = 0; i < examplestring.length(); i++) {
char c = examplestring.charAt(i);
if (i % 2 == 0) {
output += examplestring.substring(i, i + 1).toUpperCase();
} else {
output += examplestring.substring(i, i + 1);
}
}
System.out.println(output);
}
}
I have an array of string say
A=["hello", "you"]
I have a string, say
s="hello, hello you are so wonderful"
I need to count the number of occurrence of strings from A in s.
In this case, the number of occurrences is 3 (2 "hello", 1 "you").
How to do this effectively? (A might contains lots of words, and s might be long in practice)
Try:
Map<String, Integer> wordCount = new HashMap<>();
for(String a : dictionnary) {
wordCount.put(a, 0);
}
for(String s : text.split("\\s+")) {
Integer count = wordCount.get(s);
if(count != null) {
wordCount.put(s, count + 1);
}
}
public void countMatches() {
String[] A = {"hello", "you"};
String s = "hello, hello you are so wonderful";
String patternString = "(" + StringUtils.join(A, "|") + ")";
Pattern pattern = Pattern.compile(patternString);
Matcher matcher = pattern.matcher(s);
int count = 0;
while (matcher.find()) {
count++;
}
System.out.println(count);
}
Note that StringUtils is from apache commons. If you don't want to include and additional jar you can just construct that string using a for loop.
HashSet<String> searchWords = new HashSet<String>();
for(String a : dictionary) {
searchWords.add(a);
}
int count = 0;
for(String s : input.split("[ ,]")) {
if(searchWords.contains(s)) {
count++;
}
}
int count =0;
for(int i=0;i<A.length;i++)
{
count = count + s.split(A[i],-1).length - 1;
}
Working Ideone : http://ideone.com/Z9K3JX
This is fully working method with output :)
public static void main(String[] args) {
String[] A={"hello", "you"};
String s= "hello, hello you are so wonderful";
int[] count = new int[A.length];
for (int i = 0; i < A.length; i++) {
count[i] = (s.length() - s.replaceAll(A[i], "").length())/A[i].length();
}
for (int i = 0; i < count.length; i++) {
System.out.println(A[i] + ": " + count[i]);
}
}
What does this line do?
count[i] = (s.length() - s.replaceAll(A[i], "").length())/A[i].length();
This part s.replaceAll(A[i], "") changes all "hello" to empty "" string in the text.
So I take the length of everything s.length() I substract from it the length of same string without that word s.replaceAll(A[i], "").length() and I divide it by the length of that word /A[i].length()
Sample output for this example :
hello: 2
you: 1
You can use the String Tokenizer
Do something like this:
A = ["hello", "you"];
s = "hello, hello you are so wonderful";
StringTokenizer st = new StringTokenizer(s);
while (st.hasMoreElements()) {
for (String i: A) {
if(st.nextToken() == i){
//You can keep going from here
}
}
}
This is what I came up with:
It doesn't create any new objects. It uses String.indexOf(String, int), keeps track of the current index, and increments the occurance-count.
public class SearchWordCount {
public static final void main(String[] ignored) {
String[] searchWords = {"hello", "you"};
String input = "hello, hello you are so wonderful";
for(int i = 0; i < searchWords.length; i++) {
String searchWord = searchWords[i];
System.out.print(searchWord + ": ");
int foundCount = 0;
int currIdx = 0;
while(currIdx != -1) {
currIdx = input.indexOf(searchWord, currIdx);
if(currIdx != -1) {
foundCount++;
currIdx += searchWord.length();
} else {
currIdx = -1;
}
}
System.out.println(foundCount);
}
}
}
Output:
hello: 2
you: 1
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;
}
}