Develop Reference

parsing/converting task with characters and numbers within

It is necessary to repeat the character, as many times as the number behind it.
They are positive integer numbers.
case #1
input: "abc3leson11"
output: "abccclesonnnnnnnnnnn"
I already finish it in the following way:
String a = "abbc2kd3ijkl40ggg2H5uu";
String s = a + "*";
String numS = "";
int cnt = 0;
for (int i = 0; i < s.length(); i++) {
char ch = s.charAt(i);
if (Character.isDigit(ch)) {
numS = numS + ch;
cnt++;
} else {
cnt++;
try {
for (int j = 0; j < Integer.parseInt(numS); j++) {
System.out.print(s.charAt(i - cnt));
}
if (i != s.length() - 1 && !Character.isDigit(s.charAt(i + 1))) {
System.out.print(s.charAt(i));
}
} catch (Exception e) {
if (i != s.length() - 1 && !Character.isDigit(s.charAt(i + 1))) {
System.out.print(s.charAt(i));
}
}
cnt = 0;
numS = "";
}
}
But I wonder is there some better solution with less and cleaner code?

Could you take a look below? I'm using a library from StringUtils from Apache Common Utils to repeat character:
public class MicsTest {
public static void main(String[] args) {
String input = "abc3leson11";
String output = input;
Pattern p = Pattern.compile("\\d+");
Matcher m = p.matcher(input);
while (m.find()) {
int number = Integer.valueOf(m.group());
char repeatedChar = input.charAt(m.start()-1);
output = output.replaceFirst(m.group(), StringUtils.repeat(repeatedChar, number));
}
System.out.println(output);
}
}
In case you don't want to use StringUtils. You can use the below custom method to achieve the same effect:
public static String repeat(char c, int times) {
char[] chars = new char[times];
Arrays.fill(chars, c);
return new String(chars);
}

Using java basic string regx should make it more terse as follows:
public class He1 {
private static final Pattern pattern = Pattern.compile("[a-zA-Z]+(\\d+).*");
// match the number between or the last using regx;
public static void main(String... args) {
String s = "abc3leson11";
System.out.println(parse(s));
s = "abbc2kd3ijkl40ggg2H5uu";
System.out.println(parse(s));
}
private static String parse(String s) {
Matcher matcher = pattern.matcher(s);
while (matcher.find()) {
int num = Integer.valueOf(matcher.group(1));
char prev = s.charAt(s.indexOf(String.valueOf(num)) - 1);
// locate the char before the number;
String repeated = new String(new char[num-1]).replace('\0', prev);
// since the prev is not deleted, we have to decrement the repeating number by 1;
s = s.replaceFirst(String.valueOf(num), repeated);
matcher = pattern.matcher(s);
}
return s;
}
}
And the output should be:
abccclesonnnnnnnnnnn
abbcckdddijkllllllllllllllllllllllllllllllllllllllllggggHHHHHuu

String g(String a){
String result = "";
String[] array = a.split("(?<=\\D)(?=\\d)|(?<=\\d)(?=\\D)");
//System.out.println(java.util.Arrays.toString(array));
for(int i=0; i<array.length; i++){
String part = array[i];
result += part;
if(++i == array.length){
break;
}
char charToRepeat = part.charAt(part.length() - 1);
result += repeat(charToRepeat+"", new Integer(array[i]) - 1);
}
return result;
}
// In Java 11 this could be removed and replaced with the builtin `str.repeat(amount)`
String repeat(String str, int amount){
return new String(new char[amount]).replace("\0", str);
}
Try it online.
Explanation:
The split will split the letters and numbers:
abbc2kd3ijkl40ggg2H5uu would become ["abbc", "2", "kd", "3", "ijkl", "40", "ggg", "2", "H", "5", "uu"]
We then loop over the parts and add any strings as is to the result.
We then increase i by 1 first and if we're done (after the "uu") in the array above, it will break the loop.
If not the increase of i will put us at a number. So it will repeat the last character of the part x amount of times, where x is the number we found minus 1.

Here is another solution:
String str = "abbc2kd3ijkl40ggg2H5uu";
String[] part = str.split("(?<=\\d)(?=\\D)|(?=\\d)(?<=\\D)");
String res = "";
for(int i=0; i < part.length; i++){
if(i%2 == 0){
res = res + part[i];
}else {
res = res + StringUtils.repeat(part[i-1].charAt(part[i-1].length()-1),Integer.parseInt(part[i])-1);
}
}
System.out.println(res);

Yet another solution :
public static String getCustomizedString(String input) {
ArrayList<String > letters = new ArrayList<>(Arrays.asList(input.split("(\\d)")));
letters.removeAll(Arrays.asList(""));
ArrayList<String > digits = new ArrayList<>(Arrays.asList(input.split("(\\D)")));
digits.removeAll(Arrays.asList(""));
for(int i=0; i< digits.size(); i++) {
int iteration = Integer.valueOf(digits.get(i));
String letter = letters.get(i);
char c = letter.charAt(letter.length()-1);
for (int j = 0; j<iteration -1 ; j++) {
letters.set(i,letters.get(i).concat(String.valueOf(c)));
}
}
String finalResult = "";
for (String str : letters) {
finalResult += str;
}
return finalResult;
}
The usage:
public static void main(String[] args) {
String testString1 = "abbc2kd3ijkl40ggg2H5uu";
String testString2 = "abc3leson11";
System.out.println(getCustomizedString(testString1));
System.out.println(getCustomizedString(testString2));
}
And the result:
abbcckdddijkllllllllllllllllllllllllllllllllllllllllggggHHHHHuu
abccclesonnnnnnnnnnn

Extracting Ints from Strings in Java

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;
}

Convert alternate char to uppercase

I am new to java programming. I want to print a string with alternate characters in UpperCase.
String x=jTextField1.getText();
x=x.toLowerCase();
int y=x.length();
for(int i=1;i<=y;i++)
{}
I don't know how to proceed further. I want to do this question with the help of looping and continue function.
Help would be appreciated. Thanks.

#Test
public void alternateUppercase(){
String testString = "This is a !!!!! test - of the emergency? broadcast System.";
char[] arr = testString.toLowerCase().toCharArray();
boolean makeUppercase = true;
for (int i=0; i<arr.length; i++) {
if(makeUppercase && Character.isLetter(arr[i])) {
arr[i] = Character.toUpperCase(arr[i]);
makeUppercase = false;
} else if (!makeUppercase && Character.isLetter(arr[i])) {
makeUppercase = true;
}
}
String convertedString = String.valueOf(arr);
System.out.println(convertedString);
}

First, java indexes start at 0 (not 1). I think you are asking for something as simple as alternating calls to Character.toLowerCase(char) and Character.toUpperCase(char) on the result of modulo (remainder of division) 2.
String x = jTextField1.getText();
for (int i = 0, len = x.length(); i < len; i++) {
char ch = x.charAt(i);
if (i % 2 == 0) {
System.out.print(Character.toLowerCase(ch));
} else {
System.out.print(Character.toUpperCase(ch));
}
}
System.out.println();

Strings start at index 0 and finish at index x.length()-1
To look up a String by index you can use String.charAt(i)
To convert a character to upper case you can do Character.toUpperCase(ch);
I suggest you build a StringBuilder from these characters which you can toString() when you are done.

you can make it using the 65 distnace of lower case and upper case ABCabc from the unicode table like:
String str = "abbfFDdshFSDjdFDSsfsSdoi".toLowerCase();
char c;
boolean state = false;
String newStr = "";
for (int i=0; i<str.length(); i++){
c = str.charAt(o);
if (state){
newStr += c;
}
else {
newStr += c + 65;
}
state = !state;
}

I'm sure there is a slicker way to do this, but this will work for a 2 minute-answer:
public String homeWork(){
String x = "Hello World";
StringBuilder sb = new StringBuilder();
for(int i=0;i<=x.length();i++){
char c = x.charAt(i);
if(i%2==0){
sb.append(String.valueOf(c).toUpperCase());
} else {
sb.append(String.valueOf(c).toLowerCase());
}
}
return sb.toString();
}
To explain i%2==0, if the remainder of i divided by 2 is equal to zero (even numbered) return true

public class PrintingStringInAlternativeCase {
public static void main(String s[])
{
String testString = "TESTSTRING";
String output = "";
for (int i = 0; i < testString.length(); i++) {
if(i%2 == 0)
{
output += Character.toUpperCase(testString.toCharArray()[i]);
}else
{
output += Character.toLowerCase(testString.toCharArray()[i]);
}
}
System.out.println("Newly generated string is as follow: "+ output);
}
}

Using as much of your code as I could, here's what I got. First I made a string called build that will help you build your resulting string. Also, I changed the index to go from [0,size-1] not [1,size]. Using modulo devision of 2 helps with the "every other" bit.
String build =""
String x=jTextField1.getText();
x=x.toLowerCase();
int y=x.length();
for(int i=0;i<y;i++)
{
if(i%2==0){
build+=Character.toUpperCase(x.charAt(i));
else{
build+=x.charAt(i);
}
}
x=build; //not necessary, you could just use build.
Happy oding! Leave a comment if you have any questions.

public static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
System.out.println("Enter Stirng");
String str=sc.nextLine();
for(int i=0;i<str.length();i++)
{
if(i%2==0)
{
System.out.print(Character.toLowerCase(str.charAt(i)));
}
else
{
System.out.print(Character.toUpperCase(str.charAt(i)));
}
}
sc.close();
}

Java 8 Solution:
static String getMixedCase(String str) {
char[] chars = str.toCharArray();
return IntStream.range(0, str.length())
.mapToObj(i -> String.valueOf(i % 2 == 1 ? chars[i] : Character.toUpperCase(chars[i])))
.collect(Collectors.joining());
}

public class ClassC {
public static void main(String[] args) {
String str = "Hello";
StringBuffer strNew = new StringBuffer();
for (int i = 0; i < str.length(); i++) {
if (i % 2 == 0) {
strNew.append(Character.toLowerCase(str.charAt(i)));
} else {
strNew.append(Character.toUpperCase(str.charAt(i)));
}
}
System.out.println(strNew);
}
}

Capitalizing every other word in a string

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);
}
}

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;
}
}