Number of rotations required to make string

Number of rotations required to make string - java

This is my code to count the number of rotations.
But IDK, What is the problem with it.
Can anyone explain and help me out.
Test Case: Input: david vidda
Output: 2
I tried to have brute force approach but, that wasn't working even.
Can anyone point out my mistake??
import java.util.*;
class solution{
public static int arrayLeftRotation(StringBuilder str1, StringBuilder str2)
{
int i;
int count =0;
for (i = 0; i < str1.length(); i++){
if(str1.equals(str2))
{
count++;
str1 = leftRotatebyOne(str1);
System.out.println(str1);
}
else return count;
}
return count;
}
static StringBuilder leftRotatebyOne(StringBuilder str)
{
int i;
char temp = str.charAt(0);
for (i = 0; i < str.length()-1; i++)
str.setCharAt(str.indexOf(str.charAt(i)+""),str.charAt(i+1));
str.setCharAt(i,temp);
return str;
}
public static void main(String[] args)
{
Scanner in = new Scanner(System.in);
String strr1= in.nextLine();
StringBuilder str1 = new StringBuilder(strr1);
String strr2 = in.nextLine();
StringBuilder str2 = new StringBuilder(strr2);
System.out.print(arrayLeftRotation(str1, str2));
}
}

Your method leftRotateByOne appears more complicated than necessary.
Try this:
public class Solution
{
public static int arrayLeftRotation(String str1,
String str2)
{
int nr_rotate;
int counter;
nr_rotate = 0;
for (counter = 0; counter < str1.length(); counter++)
{
if (str1.equals(str2))
return (nr_rotate);
else
{
str1 = leftRotateByOne(str1);
nr_rotate++;
System.out.println(str1);
}
}
// No possible solution
return (-1);
} // arrayLeftRotation
public static String leftRotateByOne(String str)
{
return (str.substring(1) + str.charAt(0));
}
public static void main(String[] args)
{
String str1 = "david";
String str2 = "vidda";
System.out.print(arrayLeftRotation(str1, str2));
}
} // class Solution

Another possible solution for arrayLeftRotation,
public static int arrayLeftRotation(String str1, String str2) {
StringBuilder builder = new StringBuilder(str1);
for (int i = 0; i < str1.length(); i++) {
builder.append(str1.charAt(i)).delete(0, 1);
if (str2.equals(builder.toString())) {
return i + 1;
}
}
return -1;
}
Note: this will return -1 if no matches found.

The trick is to append the input string to itself, then call String#indexOf. It will give you the index at which the doubled string contains the expected string, which is what you're looking for.
Example:
public static int numberOfRotations(String input, String expected) {
final String doubledInput = input + input;
return doubledInput.indexOf(expected);
}
If you really want to implement it yourself, you need to simplify your code to minimize the possibility of making mistakes.
public static String rotate(String input) {
return input.substring(1) + input.charAt(0);
}
public static int numberOfRotations(String input, String expected) {
// handle edge cases (null, empty, etc.) here
String rotatedInput = input;
int count = 0;
while (!rotatedInput.equals(expected) && count < input.length()) {
rotatedInput = rotate(rotatedInput);
count++;
}
return count == input.length() ? -1 : count;
}

I am just trying to point out where your error lies and fix it.
Your error lies here in your leftRotatebyOne:
for (i = 0; i < str.length()-1; i++)
str.setCharAt(str.indexOf(str.charAt(i)+""),str.charAt(i+1)); // your error while shifting to the left;
What you are trying to do is shifting one position to the left, and you should just do it as:
for (i = 0; i < str.length()-1; i++)
str.setCharAt(i,str.charAt(i+1));
And then your method will work.
But I have to say Alex M has provided a cleaner solution to your problem. Perhaps you should have a try.
Your solution then can be (after the fix):
public class RotationCount {
public static int arrayLeftRotation(StringBuilder str1, StringBuilder str2) {
int i;
int count = 0;
for (i = 0; i < str1.length(); i++) {
if (!str1.toString().equals(str2.toString())) {
count++;
str1 = leftRotatebyOne(str1);
} else return count;
}
return count;
}
static StringBuilder leftRotatebyOne(StringBuilder str) {
int i;
char temp = str.charAt(0);
for (i = 0; i < str.length() - 1; i++) {
str.setCharAt(i, str.charAt(i + 1));
}
str.setCharAt(i, temp);
return str;
}
public static void main(String[] args) {
StringBuilder str1 = new StringBuilder("david");
StringBuilder str2 = new StringBuilder("vidda");
System.out.print(arrayLeftRotation(str1, str2));
}
}

Related

counter for string search in another string

I want to search how many times a string appear in another string
It does not work correctly when i put an similar string at the end.
public class C3_Project3_WPr {
public static void main(String[] args) {
String strn1="AliAliAli";
String strn2="AliAliSinaAli";
String strn3="Ali";
int count1=StringCounter(strn1, strn3);
System.out.println(count1);
int count2=StringCounter(strn2, strn3);
System.out.println(count2);
}
//ُString in String Method
static int StringCounter(String str1, String str2){
int counter=0;
if (str1.isEmpty() || str2.isEmpty()) {
return 0;
}
for(int i= 0; i<str1.length(); i++){
if(str1.contains(str2)){
counter++;
str1= str1.substring(str2.length());
}
}
return counter;
}
}

Solution to your problem is here
public static void main(String[] args) {
String strn1 = "AliAliAliwewdwdweAli";
String strn2 = "AliAliSinaAliAlAli";
String strn3 = "Ali";
int count1 = StringCounter(strn1, strn3);
System.out.println(count1);
int count2 = StringCounter(strn2, strn3);
System.out.println(count2);
}
// ُString in String Method
static int StringCounter(String str1, String str2) {
int counter = 0;
if (str1.isEmpty() || str2.isEmpty()) {
return 0;
}
for (int i = str1.indexOf(str2); i >= 0; i = str1.indexOf(str2, i + str2.length())) {
counter++;
}
return counter;
}
}

When modifying str1 you only take in to account the length of the search string, but ignore the index in which it was found. Fixing this (e.g., by using indexOf), will fix your results too:
int index = str1.indexOf(str2);
while (index >= 0) {
counter++;
index = str1.indexOf(str2, index + str2.length());
}

Use recursive method: It's quick and easy way to solve your problem:
static int StringCounter(String str1, String str2){
return (str1.contains(str2)) ? 1 + StringCounter(str1.replaceFirst(str2, ""), str2) : 0;
}

Java Compress no return?

what my code is supposed to be doing is converting input strings and outputing the compressed versions
Ex. Input: "qqqwww" Output: "3q3w".
But my code returns nothing.
P.S. IO is just an input system.
public class Compress {
public static String compress(String original){
String s = "";
char s1 = original.charAt(0);
int count = 1;
for(int i = 0; i < original.length(); i++){
char c = original.charAt(i);
if(c == s1){
count++;
}
else{
s = s + count + s1; //i think the problem is here right???
count = 1;
}
s1 = c;
}
return s;
}
public static void main(String[] args){
String s = IO.readString();
String y = compress(s);
System.out.println(y);
}
}

Your could should look like this :
String returnString="";
for (int index = 0; index < original.length();) {
char currentChar = original.charAt(index);
int counter=1;
while(++index < original.length() && currentChar==original.charAt(index)) {
counter++;
}
returnString=returnString+counter+currentChar;
}
return returnString;
}
Here we loop thought the string (outer for loop) and check if the adjacent values are same the we keep adding them. (inner while loop)

Converting a number to binary java code

I am writing a code to convert a number to binary representation.Here's my code.It doesn't give the correct answer but I can't figure out where I am making the mistake.If someone could point out where my mistake and how to correct it I would be grateful.
public class ConvertBinary{
public static String convertBinary(int n){
String s="";
while(n>0){
s+=(n%2);
n=n/2;
}
int len=s.length();
String[] binary=s.split("");
String[] binaryCopy=new String[s.length()];
for(int i=0;i<len;i++){
binaryCopy[i]=binary[len-i-1];
s+=binaryCopy[i];
}
return s;
}
public static void main (String args[]){
System.out.println(convertBinary(19));
}
}

Apart from all these answers the problem with your code was that you are not clearing the string before reversing it. So before the for loop just put s = "" and you code should work fine.. :)
Based on comment
public class ConvertBinary {
public static String convertBinary(int n) {
String s = "";
while (n > 0) {
s += (n % 2);
n = n / 2;
}
int len = s.length();
String[] binary = s.split("");
String[] binaryCopy = new String[s.length()];
s = "";
for (int i = 0; i < len; i++) {
binaryCopy[i] = binary[len - i - 1];
s += binaryCopy[i];
}
return s;
}
public static void main(String args[]) {
int num = 4;
System.out.println(convertBinary(num));
System.out.println(Integer.toBinaryString(num));
}
}

public static String convertBinary(int n){
String s="";
while(n>0){
s+=(n%2);
n=n/2;
}
return (new StringBuffer(s).reverse().toString());
}

If you're looking for error in your implementation, you'd rather put:
s = (n % 2) + s;
isntead of
s+=(n%2);
so the code'll be
// n should be positive
public static String convertBinary(int n){
if (n == 0)
return "0";
String s = "";
// for is more compact than while here
for (; n > 0; n /= 2)
s = (n % 2) + s;
return s;
}
however in real life
Integer.toString(n, 2);
is much more convenient

Use Java standard library: http://docs.oracle.com/javase/7/docs/api/java/lang/Integer.html#toString%28int,%20int%29
public class ConvertBinary{
public static String convertBinary(int n){
return Integer.toString(n, 2);
}
public static void main (String args[]){
System.out.println(ConveryBinary.convertBinary(19));
}
}
EDIT: As #Holger says, there's also a toBinaryString: http://docs.oracle.com/javase/7/docs/api/java/lang/Integer.html#toBinaryString%28int%29
public static String convertBinary(int n){
return Integer.toBinaryString(n);
}

How to Find the Longest Palindrome (java)

Hi I've been doing this java program, i should input a string and output the longest palindrome that can be found ..
but my program only output the first letter of the longest palindrome .. i badly need your help .. thanks!
SHOULD BE:
INPUT : abcdcbbcdeedcba
OUTPUT : bcdeedcb
There are two palindrome strings : bcdcb and bcdeedcb
BUT WHEN I INPUT : abcdcbbcdeedcba
output : b
import javax.swing.JOptionPane;
public class Palindrome5
{ public static void main(String args[])
{ String word = JOptionPane.showInputDialog(null, "Input String : ", "INPUT", JOptionPane.QUESTION_MESSAGE);
String subword = "";
String revword = "";
String Out = "";
int size = word.length();
boolean c;
for(int x=0; x<size; x++)
{ for(int y=x+1; y<size-x; y++)
{ subword = word.substring(x,y);
c = comparisonOfreverseword(subword);
if(c==true)
{
Out = GetLongest(subword);
}
}
}
JOptionPane.showMessageDialog(null, "Longest Palindrome : " + Out, "OUTPUT", JOptionPane.PLAIN_MESSAGE);
}
public static boolean comparisonOfreverseword(String a)
{ String rev = "";
int tempo = a.length();
boolean z=false;
for(int i = tempo-1; i>=0; i--)
{
char let = a.charAt(i);
rev = rev + let;
}
if(a.equalsIgnoreCase(rev))
{
z=true;
}
return(z);
}
public static String GetLongest(String sWord)
{
int sLength = sWord.length();
String Lpalindrome = "";
int storage = 0;
if(storage<sLength)
{
storage = sLength;
Lpalindrome = sWord;
}
return(Lpalindrome);
}
}

modified program..this program will give the correct output
package pract1;
import javax.swing.JOptionPane;
public class Palindrome5
{
public static void main(String args[])
{
String word = JOptionPane.showInputDialog(null, "Input String : ", "INPUT", JOptionPane.QUESTION_MESSAGE);
String subword = "";
String revword = "";
String Out = "";
int size = word.length();
boolean c;
String Lpalindrome = "";
int storage=0;
String out="";
for(int x=0; x<size; x++)
{ for(int y=x+1; y<=size; y++)
{ subword = word.substring(x,y);
c = comparisonOfreverseword(subword);
if(c==true)
{
int sLength = subword.length();
if(storage<sLength)
{
storage = sLength;
Lpalindrome = subword;
out=Lpalindrome;
}
}
}
}
JOptionPane.showMessageDialog(null, "Longest Palindrome : " + out, "OUTPUT", JOptionPane.PLAIN_MESSAGE);
}
public static boolean comparisonOfreverseword(String a)
{ String rev = "";
int tempo = a.length();
boolean z=false;
for(int i = tempo-1; i>=0; i--)
{
char let = a.charAt(i);
rev = rev + let;
}
if(a.equalsIgnoreCase(rev))
{
z=true;
}
return(z);
}
}

You have two bugs:
1.
for(int y=x+1; y<size-x; y++)
should be
for(int y=x+1; y<size; y++)
since you still want to go all the way to the end of the string. With the previous loop, since x increases throughout the loop, your substring sizes decrease throughout the loop (by removing x characters from their end).
2.
You aren't storing the longest string you've found so far or its length. The code
int storage = 0;
if(storage<sLength) {
storage = sLength;
...
is saying 'if the new string is longer than zero characters, then I will assume it is the longest string found so far and return it as LPalindrome'. That's no help, since we may have previously found a longer palindrome.
If it were me, I would make a static variable (e.g. longestSoFar) to hold the longest palindrome found so far (initially empty). With each new palindrome, check if the new one is longer than longestSoFar. If it is longer, assign it to longestSoFar. Then at the end, display longestSoFar.
In general, if you're having trouble 'remembering' something in the program (e.g. previously seen values) you have to consider storing something statically, since local variables are forgotten once their methods finish.

public class LongestPalindrome {
/**
* #param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
String S= "abcdcba";
printLongestPalindrome(S);
}
public static void printLongestPalindrome(String S)
{
int maxBack=-1;
int maxFront = -1;
int maxLength=0;
for (int potentialCenter = 0 ; potentialCenter < S.length();potentialCenter ++ )
{
int back = potentialCenter-1;
int front = potentialCenter + 1;
int longestPalindrome = 0;
while(back >=0 && front<S.length() && S.charAt(back)==S.charAt(front))
{
back--;
front++;
longestPalindrome++;
}
if (longestPalindrome > maxLength)
{
maxLength = longestPalindrome+1;
maxBack = back + 1;
maxFront = front;
}
back = potentialCenter;
front = potentialCenter + 1;
longestPalindrome=0;
while(back >=0 && front<S.length() && S.charAt(back)==S.charAt(front))
{
back--;
front++;
longestPalindrome++;
}
if (longestPalindrome > maxLength)
{
maxLength = longestPalindrome;
maxBack = back + 1;
maxFront = front;
}
}
if (maxLength == 0) System.out.println("There is no Palindrome in the given String");
else{
System.out.println("The Longest Palindrome is " + S.substring(maxBack,maxFront) + "of " + maxLength);
}
}
}

I have my own way to get longest palindrome in a random word. check this out
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
System.out.println(longestPalSubstr(in.nextLine().toLowerCase()));
}
static String longestPalSubstr(String str) {
char [] input = str.toCharArray();
Set<CharSequence> out = new HashSet<CharSequence>();
int n1 = str.length()-1;
for(int a=0;a<=n1;a++)
{
for(int m=n1;m>a;m--)
{
if(input[a]==input[m])
{
String nw = "",nw2="";
for (int y=a;y<=m;y++)
{
nw=nw+input[y];
}
for (int t=m;t>=a;t--)
{
nw2=nw2+input[t];
}
if(nw2.equals(nw))
{
out.add(nw);
break;
}
}
}
}
int a = out.size();
int maxpos=0;
int max=0;
Object [] s = out.toArray();
for(int q=0;q<a;q++)
{
if(max<s[q].toString().length())
{
max=s[q].toString().length();
maxpos=q;
}
}
String output = "longest palindrome is : "+s[maxpos].toString()+" and the lengths is : "+ max;
return output;
}
this method will return the max length palindrome and the length of it. its a way that i tried and got the answer. and this method will run whether its a odd length or even length.

public class LongestPalindrome {
public static void main(String[] args) {
HashMap<String, Integer> result = findLongestPalindrome("ayrgabcdeedcbaghihg123444456776");
result.forEach((k, v) -> System.out.println("String:" + k + " Value:" + v));
}
private static HashMap<String, Integer> findLongestPalindrome(String str) {
int i = 0;
HashMap<String, Integer> map = new HashMap<String, Integer>();
while (i < str.length()) {
String alpha = String.valueOf(str.charAt(i));
if (str.indexOf(str.charAt(i)) != str.lastIndexOf(str.charAt(i))) {
String pali = str.substring(i, str.lastIndexOf(str.charAt(i)) + 1);
if (isPalindrome(pali)) {
map.put(pali, pali.length());
i = str.lastIndexOf(str.charAt(i));
}
}
i++;
}
return map;
}
public static boolean isPalindrome(String input) {
for (int i = 0; i <= input.length() / 2; i++) {
if (input.charAt(i) != input.charAt(input.length() - 1 - i)) {
return false;
}
}
return true;
}
}
This approach is simple.
Output:
String:abcdeedcba Value:10
String:4444 Value:4
String:6776 Value:4
String:ghihg Value:5

This is my own way to get longest palindrome. this will return the length and the palindrome word
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
System.out.println(longestPalSubstr(in.nextLine().toLowerCase()));
}
static String longestPalSubstr(String str) {
char [] input = str.toCharArray();
Set<CharSequence> out = new HashSet<CharSequence>();
int n1 = str.length()-1;
for(int a=0;a<=n1;a++)
{
for(int m=n1;m>a;m--)
{
if(input[a]==input[m])
{
String nw = "",nw2="";
for (int y=a;y<=m;y++)
{
nw=nw+input[y];
}
for (int t=m;t>=a;t--)
{
nw2=nw2+input[t];
}
if(nw2.equals(nw))
{
out.add(nw);
break;
}
}
}
}
int a = out.size();
int maxpos=0;
int max=0;
Object [] s = out.toArray();
for(int q=0;q<a;q++)
{
if(max<s[q].toString().length())
{
max=s[q].toString().length();
maxpos=q;
}
}
String output = "longest palindrome is : "+s[maxpos].toString()+" and the lengths is : "+ max;
return output;
}
this method will return the max length palindrome and the length of it. its a way that i tried and got the answer. and this method will run whether its a odd length or even length.

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;
}
}

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