I am trying to write a program that will allows users to make short blog entries by typing abbreviations for common words. On completion of the input, Program will expand the abbreviations according to the lexicon defined.
Conditions
A substituted word must be the shortest word that can be formed by adding zero or more letters (or punctuation symbols) to the abbreviation.
If two or more unique words can be formed by adding the same number of letters, then the abbreviation should be printed as it is.
Input
The input is divided into two sections.
The first section is the lexicon itself, and the second section is a user's blog entry that needs to be expanded. The sections are divided by a single | character.
For example:-
cream chocolate every ever does do ice is fried friend friends lick like floor favor flavor flower best but probably poorly say says that what white our you your strawberry storyboard the | wht flvr ic crm ds yr bst fnd lke? ur frds lk stbry, bt choc s prly th bs flr vr!
Output
what flavor ice cream does your best friend like? our friends lk strawberry, but chocolate is poorly the best floor ever!
I have written the program for this and tested it locally with many different test cases with success but it fails on submission to test server.
An automated Test suit runs to validate the program’s output on its submission to test server. In case of failure, details of the failing test case/cases are not visible.
Below is the program
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class BlogEntry {
/**
* #param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
String[][] info = readInput();
String[] output = inputExpander(info[0],info[1]);
//System.out.println();
for(int i = 0; i < output.length; ++i) {
if(i!=0)
System.out.print(" ");
System.out.print(output[i]);
}
}
public static String[][] readInput() {
BufferedReader bufferReader = new BufferedReader(new InputStreamReader(
System.in));
String input = null;
String[][] info = new String[2][];
String[] text;
String[] abbr;
try {
input = bufferReader.readLine();
StringTokenizer st1 = new StringTokenizer(input, "|");
String first = "", second = "";
int count = 0;
while (st1.hasMoreTokens()) {
++count;
if(count == 1)
first = st1.nextToken();
if(count == 2)
second = st1.nextToken();
}
st1 = new StringTokenizer(first, " ");
count = st1.countTokens();
text = new String[count];
count = 0;
while (st1.hasMoreTokens()) {
text[count] = st1.nextToken();
count++;
}
st1 = new StringTokenizer(second, " ");
count = st1.countTokens();
abbr = new String[count];
count = 0;
while (st1.hasMoreTokens()) {
abbr[count] = st1.nextToken();
count++;
}
info[0] = text;
info[1] = abbr;
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
return info;
}
public static String[] inputExpander(String[] text, String[] abbr) {
String[] output = new String[abbr.length];
boolean result;
for (int i = 0; i < abbr.length; ++i) {
String abbrToken = abbr[i];
char[] char_abbr_token = abbrToken.toCharArray();
for (int j = 0; j < text.length; ++j) {
String textToken = text[j];
boolean flag2 = false;
if ((char_abbr_token[char_abbr_token.length - 1] == '!')
|| (char_abbr_token[char_abbr_token.length - 1] == '?')
|| (char_abbr_token[char_abbr_token.length - 1] == ',')
|| (char_abbr_token[char_abbr_token.length - 1] == ';')) {
flag2 = true;
}
char[] char_text_token = textToken.toCharArray();
result = ifcontains(char_text_token, char_abbr_token);
if (result) {
int currentCount = textToken.length();
int alreadyStoredCount = 0;
if (flag2)
textToken = textToken
+ char_abbr_token[char_abbr_token.length - 1];
if (output[i] == null)
output[i] = textToken;
else {
alreadyStoredCount = output[i].length();
char[] char_stored_token = output[i].toCharArray();
if ((char_stored_token[char_stored_token.length - 1] == '!')
|| (char_stored_token[char_stored_token.length - 1] == '?')
|| (char_stored_token[char_stored_token.length - 1] == ',')
|| (char_stored_token[char_stored_token.length - 1] == ';')) {
alreadyStoredCount -= 1;
}
if (alreadyStoredCount > currentCount) {
output[i] = textToken;
} else if (alreadyStoredCount == currentCount) {
output[i] = abbrToken;
}
}
}
}
if(output[i] == null)
output[i] = abbrToken;
}
return output;
}
public static boolean ifcontains(char[] char_text_token,
char[] char_abbr_token) {
int j = 0;
boolean flag = false;
for (int i = 0; i < char_abbr_token.length; ++i) {
flag = false;
for (; j < char_text_token.length; ++j) {
if ((char_abbr_token[i] == '!') || (char_abbr_token[i] == '?')
|| (char_abbr_token[i] == ',')
|| (char_abbr_token[i] == ';')) {
flag = true;
break;
}
if (char_abbr_token[i] == char_text_token[j]) {
flag = true;
break;
}
}
if (!flag)
return flag;
}
//System.out.println("match found" + flag);
return flag;
}
}
Can someone direct/hint me to/about the possible use case which I may have missed in the implementation? Thanks in advance.
Ran your program with duplicate word in input (lexicon). When a word is repeated in the lexicon, it is not getting expanded because the check is only on the length(line no. 112) of the stored word not its content.
I think you need to check:-
If same word appears more than once then expand.
If 2 or more unique words of same length appear then keep it short.
How would I approach solving this:
Parse the input, tokenize the lexicon and the text.
For each (possibly abbreviated) token like choc convert it to a regular expression like .*c.*h.*o.*c.*.
Search for shortest lexicon words matching this regular expression. Replace the text token if exactly one is found, otherwise leave it alone.
It is quite hard to say what's wrong with your code without careful debugging. It is hard to understand what one or the other part of the code does, it's not quite self-evident.
Related
I've recently started java and I want to compress a string like this:
Input:aaaaabbbbwwwccc Output:a5b4w3c3
Input:aaabbccds Output:a3b2c2ds
Input:Abcd Output:Abcd
The following code is what I have done but, it does not work.
public class CompressString {
public static void main(String[] args) {
String out = "";
Scanner in = new Scanner(System.in);
String input = in.next();
int length = input.length();
int counter = 1;
if (length == 0) {
System.out.println(" ");
} else {
for (int i = 0; i<length;i++){
if (input.charAt(i)==input.charAt(i+1)){
counter++;
}else {
if (counter == 1){
out = out+input.charAt(i-counter);
}else{
out = out+input.charAt(i-counter)+counter;
}
}
i++;
counter = 1;
}
System.out.println(out.toString());
}
}
}
The simplest program to do that would loop through each character in the string and check when the character is different from the previous seen one and, if so, add the last one and its count to the compressed string:
String input = "aaaaabbbbwwwccc";
StringBuilder compressed = new StringBuilder();
char last = 0;
int lastCount = 0;
for (int i = 0; i < input.length(); i++) {
char c = input.charAt(i);
if (last == 0 || c != last) {
if (lastCount != 0) {
compressed.append(last);
if (lastCount > 1) {
compressed.append(lastCount);
}
}
last = c;
lastCount = 1;
} else {
lastCount++;
}
}
// take care of the last repeating sequence if any
if (lastCount > 0) {
compressed.append(last);
if (lastCount > 1) {
compressed.append(lastCount);
}
}
Here is a very compact way of doing this with a regex matcher along with a string buffer:
String input = "aaaaabbbbwwwccc";
Pattern r = Pattern.compile("(.)\\1{0,}");
Matcher m = r.matcher(input);
StringBuffer buffer = new StringBuffer();
while (m.find()) {
m.appendReplacement(buffer, m.group(1) + m.group(0).length());
}
m.appendTail(buffer);
System.out.println(buffer.toString());
This prints:
a5b4w3c3
For an explanation, the above logic searches for the regex pattern (.)\1{0,}. This will match any single character, along with that same character occurring again possibly one or more times afterwards. It then replaces with just the single character followed by the count of the number of times it occurs.
I'm working on a project where I want to be able to be able to parse some text and find nouns and a lot of the text I want to parse has pronouns in it for Example => "Emma the parrot was a bird. She lived in a tall tree".
I don't want to work with "She's" etc. as they aren't seen as nouns in the dictionary I'm working with so I've been working on a method to replace She etc with the previous occurrence of a name. So the above example would output to => "Emma the parrot was a bird. Emma lived in a tall tree".
The method is working fine when I have a small sample however when I'm working with 3-4 different people in one text it doesn't work.
public static String replacePronouns(String text, ArrayList<String> dictionary) {
String[] strArray = text.replaceAll("\\.", " .").replaceAll("\\,", "").split("\\s+");
String previousName = "";
for(int i = 0; i < strArray.length; i++ ) {
//we'll have to set this to be more dynamic -> change to pronouns in dicitonary
if(strArray[i].equals("His") || strArray[i].equals("She") || strArray[i].equals("she") || strArray[i].equals("him") || strArray[i].equals("he") || strArray[i].equals("her")) {
for(int j = (i-1); j>=0; j--) {
int count = dictionary.size()-1;
boolean flag = false;
while(count>=0 && flag==false) {
if(strArray[j].equals(dictionary.get(count).split(": ")[1]) && dictionary.get(count).split(": ")[0].equals("Name")) {
previousName = strArray[j];
flag = true; }
count--;
} }
strArray[i] = previousName; } }
return Arrays.toString(strArray).replaceAll("\\[", "").replaceAll("\\,", "").replaceAll("\\]", "");
}
It takes in my text
String text = "Karla was a bird and she had beautifully colorful feathers. She lived in a tall tree.
And a "dictionary"
ArrayList<String> dictionary = new ArrayList<>();
dictionary.add("Name: hunter");
dictionary.add("Name: Karla");
dictionary.add("Noun: hawk");
dictionary.add("Noun: feathers");
dictionary.add("Noun: tree");
dictionary.add("Noun: arrows");
dictionary.add("Verb: was a");
dictionary.add("Verb: had");
dictionary.add("Verb: missed");
dictionary.add("Verb: knew");
dictionary.add("Verb: offered");
dictionary.add("Verb: pledged");
dictionary.add("Verb: shoot");
But it always outputs Karla in this example, even if we had "The hunter shot his gun" in the same string.
Any help on why this isn't working would be appreciated
This isn't working because you continue looping over j even after you've found a match in the dictionary. That is - you keep looking back towards the beginning of the string, and eventually find "Karla", even though you've already matched "hunter".
There are many ways you could fix this. One very simple one would be to move boolean flag = false; up to before the for loop over j, and change the condition from j >= 0 to j >= 0 && !flag, so that you stop looping as soon as flag is true. Like so :
public static String replacePronouns(String text, ArrayList<String> dictionary) {
String[] strArray = text.replaceAll("\\.", " .").replaceAll("\\,", "").split("\\s+");
String previousName = "";
for (int i = 0; i < strArray.length; i++) {
boolean flag = false;
// we'll have to set this to be more dynamic -> change to pronouns in dicitonary
if (strArray[i].equals("His") || strArray[i].equals("She") || strArray[i].equals("she") || strArray[i].equals("him") || strArray[i].equals("he") || strArray[i].equals("her")) {
for (int j = (i - 1); j >= 0 && flag == false; j--) {
int count = dictionary.size() - 1;
while (count >= 0) {
if (strArray[j].equals(dictionary.get(count).split(": ")[1]) && dictionary.get(count).split(": ")[0].equals("Name")) {
previousName = strArray[j];
flag = true;
}
count--;
}
}
strArray[i] = previousName;
}
}
return Arrays.toString(strArray).replaceAll("\\[", "").replaceAll("\\,", "").replaceAll("\\]", "");
}
If you placed your } characters in a more standard way, this kind of error would be easier to see.
I'm writing a calculator code that solves the input whats given in string. All is good, except when it gets a negative result in the parentheses it fails badly because two operations get next to each other:
1+2*(10-11) >> 1+2*(-1) >> 1+2*-1
So where *- is, it gets "" (nothing) in the BigDecimal's constructor.
I know what's the problem, but how can I solve it?
import java.math.BigDecimal;
import java.util.ArrayList;
public class DoMath {
public static void main(String[] args) {
// Test equation goes here.
String number = "95.3+43.23*(10-11.1)";
System.out.println(doMath(number));
}
public static BigDecimal doMath(String input) {
StringBuilder builtInput = new StringBuilder(input);
StringBuilder help = new StringBuilder();
// Check if there are parenthesis in the equation.
boolean noParenthesis = true;
for (int i = 0; i < builtInput.length(); i++) {
if (builtInput.charAt(i) == 40) {
noParenthesis = false;
break;
}
}
if (noParenthesis) { // If there are no parenthesis, calculate the equation!
return calculateAndConvert(builtInput);
} else { // If there are parenthesis, breakdown to simple equations!
int parenthesePair = 0;
// Start extracting characters from the builtInput variable.
for (int i = 0; i < builtInput.length(); i++) {
// Start where we find a parentheses opener.
if (builtInput.charAt(i) == 40) {
parenthesePair = 1;
builtInput.deleteCharAt(i);
for (int j = i; j < builtInput.length(); j++) {
// If we find another opener, add one to parenthesePair variable.
if (builtInput.charAt(j) == 40) {
parenthesePair++;
}
// If we find a closer, subtract one from the given variable.
if (builtInput.charAt(j) == 41) {
parenthesePair--;
}
// If we have found the matching pair, delete it and break the for loop.
if (parenthesePair == 0) {
builtInput.deleteCharAt(j);
builtInput.insert(j, doMath(help.toString()));
break;
}
help.append(builtInput.charAt(j));
builtInput.deleteCharAt(j);
j--;
}
break;
}
}
}
System.out.println(builtInput);
return doMath(builtInput.toString());
}
public static BigDecimal calculateAndConvert(StringBuilder input) {
ArrayList<BigDecimal> listOfNumbers = new ArrayList<BigDecimal>();
StringBuilder numBay = new StringBuilder();
StringBuilder operations = new StringBuilder();
// If the first character is -, the first number is negative.
boolean firstIsNegative = false;
if (input.charAt(0) == 45) {
firstIsNegative = true;
input.deleteCharAt(0);
}
// Converting to numbers.
while (input.length() != 0) {
// If the character is a number or a dot, put it in the numBay variable and delete the char.
if (input.charAt(0) >= 48 && input.charAt(0) <= 57 || input.charAt(0) == 46) {
numBay.append(input.charAt(0));
// If the character is not a number, put it in the operations variable
// and save the number in the list (not operator characters are filtered)
} else {
listOfNumbers.add(new BigDecimal(numBay.toString()));
numBay.setLength(0);
operations.append(input.charAt(0));
}
// Delete the character.
input.deleteCharAt(0);
}
listOfNumbers.add(new BigDecimal(numBay.toString()));
// Setting first number to negative if it's needed.
if (firstIsNegative) {
listOfNumbers.set(0, listOfNumbers.get(0).negate());
}
// Calculate the result from the list and operations and return it.
return calculate(listOfNumbers, operations);
}
public static BigDecimal calculate(ArrayList<BigDecimal> list, StringBuilder ops) {
BigDecimal momentaryResult;
// Check for a multiply operation - if there is one, solve it.
for (int i = 0; i < ops.length(); i++) {
if (ops.charAt(i) == 42) {
momentaryResult = list.get(i).multiply(list.get(i + 1));
list.remove(i);
list.set(i, momentaryResult);
ops.deleteCharAt(i);
i--;
}
}
// Check for a divide operation - if there is one, solve it.
for (int i = 0; i < ops.length(); i++) {
if (ops.charAt(i) == 47) {
momentaryResult = list.get(i).divide(list.get(i + 1));
list.remove(i);
list.set(i, momentaryResult);
ops.deleteCharAt(i);
i--;
}
}
// Check for a subtract operation - if there is one, solve it.
for (int i = 0; i < ops.length(); i++) {
if (ops.charAt(i) == 45) {
momentaryResult = list.get(i).subtract(list.get(i + 1));
list.remove(i);
list.set(i, momentaryResult);
ops.deleteCharAt(i);
i--;
}
}
// Check for a plus operation - if there is one, solve it.
for (int i = 0; i < ops.length(); i++) {
if (ops.charAt(i) == 43) {
momentaryResult = list.get(i).add(list.get(i + 1));
list.remove(i);
list.set(i, momentaryResult);
ops.deleteCharAt(i);
i--;
}
}
// Return with the one remaining number that represents the result.
return list.get(0);
}
}
Edit: or would it be easier to write a new code with a different algorithm...?
I would post this as a comment to your question, but I do not have the required reputation to do so.
Anyway, since you have already recognized that the bug is the "operator" *- couldn't you make a method that would fix this problem by replacing the plus operator immediately before by a minus? Like this:
1+2*-1 >>> 1-2*1
If you want I can write you the code. But maybe it will be easier for you to adapt a solution like this in your code that is already working.
Edit - 1:
Obviously, the code should also treat the following cases:
1-2*-1 >>> 1+2*1
2*-1 >>> -2*1
Edit - 2:
Here is the code I managed to make. Let me know if you find any errors.
public int countChar(String str, char chr) {
int count = 0;
for (int k = 0; k < str.length(); k++) {
if (str.charAt(k) == chr)
count++;
}
return count;
}
public String fixBug(String eq) {
boolean hasBug = eq.contains("*-");
if (hasBug) {
String subeq;
int indbug, indp, indm;
eq = eq.replace("*-", "#");
int N = countChar(eq, '#');
for (int k = N; k > 0; k--) {
indbug = eq.indexOf('#');
subeq = eq.substring(0, indbug);
indp = subeq.lastIndexOf('+');
indm = subeq.lastIndexOf('-');
if (indp == -1 && indm == -1) {
eq = "-" + eq;
} else if (indp > indm) {
eq = eq.substring(0, indp) + '-' + eq.substring(indp + 1);
} else {
eq = eq.substring(0, indm) + '+' + eq.substring(indm + 1);
}
}
eq = eq.replace("#", "*");
}
return eq;
}
Please help me to identify my mistakes in this code. I am new to Java. Excuse me if I have done any mistake. This is one of codingbat java questions. I am getting Timed Out error message for some inputs like "xxxyakyyyakzzz". For some inputs like "yakpak" and "pakyak" this code is working fine.
Question:
Suppose the string "yak" is unlucky. Given a string, return a version where all the "yak" are removed, but the "a" can be any char. The "yak" strings will not overlap.
public String stringYak(String str) {
String result = "";
int yakIndex = str.indexOf("yak");
if (yakIndex == -1)
return str; //there is no yak
//there is at least one yak
//if there are yaks store their indexes in the arraylist
ArrayList<Integer> yakArray = new ArrayList<Integer>();
int length = str.length();
yakIndex = 0;
while (yakIndex < length - 3) {
yakIndex = str.indexOf("yak", yakIndex);
yakArray.add(yakIndex);
yakIndex += 3;
}//all the yak indexes are stored in the arraylist
//iterate through the arraylist. skip the yaks and get non-yak substrings
for(int i = 0; i < length; i++) {
if (yakArray.contains(i))
i = i + 2;
else
result = result + str.charAt(i);
}
return result;
}
Shouldn't you be looking for any three character sequence starting with a 'y' and ending with a 'k'? Like so?
public static String stringYak(String str) {
char[] chars = (str != null) ? str.toCharArray()
: new char[] {};
StringBuilder sb = new StringBuilder();
for (int i = 0; i < chars.length; i++) {
if (chars[i] == 'y' && chars[i + 2] == 'k') { // if we have 'y' and two away is 'k'
// then it's unlucky...
i += 2;
continue; //skip the statement sb.append
} //do not append any pattern like y1k or yak etc
sb.append(chars[i]);
}
return sb.toString();
}
public static void main(String[] args) {
System.out.println(stringYak("1yik2yak3yuk4")); // Remove the "unlucky" strings
// The result will be 1234.
}
It looks like your programming assignment. You need to use regular expressions.
Look at http://www.vogella.com/articles/JavaRegularExpressions/article.html#regex for more information.
Remember, that you can not use contains. Your code maybe something like
result = str.removeall("y\wk")
you can try this
public static String stringYak(String str) {
for (int i = 0; i < str.length(); i++) {
if(str.charAt(i)=='y'){
str=str.replace("yak", "");
}
}
return str;
}
My goal is to write a program that compresses a string, for example:
input: hellooopppppp!
output:he2l3o6p!
Here is the code I have so far, but there are errors.
When I have the input: hellooo
my code outputs: hel2l3o
instead of: he213o
the 2 is being printed in the wrong spot, but I cannot figure out how to fix this.
Also, with an input of: hello
my code outputs: hel2l
instead of: he2lo
It skips the last letter in this case all together, and the 2 is also in the wrong place, an error from my first example.
Any help is much appreciated. Thanks so much!
public class compressionTime
{
public static void main(String [] args)
{
System.out.println ("Enter a string");
//read in user input
String userString = IO.readString();
//store length of string
int length = userString.length();
System.out.println(length);
int count;
String result = "";
for (int i=1; i<=length; i++)
{
char a = userString.charAt(i-1);
count = 1;
if (i-2 >= 0)
{
while (i<=length && userString.charAt(i-1) == userString.charAt(i-2))
{
count++;
i++;
}
System.out.print(count);
}
if (count==1)
result = result.concat(Character.toString(a));
else
result = result.concat(Integer.toString(count).concat(Character.toString(a)));
}
IO.outputStringAnswer(result);
}
}
I would
count from 0 as that is how indexes work in Java. Your code will be simpler.
would compare the current char to the next one. This will avoid printing the first character.
wouldn't compress ll as 2l as it is no smaller. Only sequences of at least 3 will help.
try to detect if a number 3 to 9 has been used and at least print an error.
use the debugger to step through the code to understand what it is doing and why it doesn't do what you think it should.
I am doing it this way. Very simple:
public static void compressString (String string) {
StringBuffer stringBuffer = new StringBuffer();
for (int i = 0; i < string.length(); i++) {
int count = 1;
while (i + 1 < string.length()
&& string.charAt(i) == string.charAt(i + 1)) {
count++;
i++;
}
if (count > 1) {
stringBuffer.append(count);
}
stringBuffer.append(string.charAt(i));
}
System.out.println("Compressed string: " + stringBuffer);
}
You can accomplish this using a nested for loops and do something simial to:
count = 0;
String results = "";
for(int i=0;i<userString.length();){
char begin = userString.charAt(i);
//System.out.println("begin is: "+begin);
for(int j=i+1; j<userString.length();j++){
char next = userString.charAt(j);
//System.out.println("next is: "+next);
if(begin == next){
count++;
}
else{
System.out.println("Breaking");
break;
}
}
i+= count+1;
if(count>0){
String add = begin + "";
int tempcount = count +1;
results+= tempcount + add;
}
else{
results+= begin;
}
count=0;
}
System.out.println(results);
I tested this output with Hello and the result was He2lo
also tested with hellooopppppp result he2l3o6p
If you don't understand how this works, you should learn regular expressions.
public String rleEncodeString(String in) {
StringBuilder out = new StringBuilder();
Pattern p = Pattern.compile("((\\w)\\2*)");
Matcher m = p.matcher(in);
while(m.find()) {
if(m.group(1).length() > 1) {
out.append(m.group(1).length());
}
out.append(m.group(2));
}
return out.toString();
}
Try something like this:
public static void main(String[] args) {
System.out.println("Enter a string:");
Scanner IO = new Scanner(System.in);
// read in user input
String userString = IO.nextLine() + "-";
int length = userString.length();
int count = 0;
String result = "";
char new_char;
for (int i = 0; i < length; i++) {
new_char = userString.charAt(i);
count++;
if (new_char != userString.charAt(i + 1)) {
if (count != 1) {
result = result.concat(Integer.toString(count + 1));
}
result = result.concat(Character.toString(new_char));
count = 0;
}
if (userString.charAt(i + 1) == '-')
break;
}
System.out.println(result);
}
The problem is that your code checks if the previous letter, not the next, is the same as the current.
Your for loops basically goes through each letter in the string, and if it is the same as the previous letter, it figures out how many of that letter there is and puts that number into the result string. However, for a word like "hello", it will check 'e' and 'l' (and notice that they are preceded by 'h' and 'e', receptively) and think that there is no repeat. It will then get to the next 'l', and then see that it is the same as the previous letter. It will put '2' in the result, but too late, resulting in "hel2l" instead of "he2lo".
To clean up and fix your code, I recommend the following to replace your for loop:
int count = 1;
String result = "";
for(int i=0;i<length;i++) {
if(i < userString.length()-1 && userString.charAt(i) == userString.charAt(i+1))
count++;
else {
if(count == 1)
result += userString.charAt(i);
else {
result = result + count + userString.charAt(i);
count = 1;
}
}
}
Comment if you need me to explain some of the changes. Some are necessary, others optional.
Here is the solution for the problem with better time complexity:
public static void compressString (String string) {
LinkedHashSet<String> charMap = new LinkedHashSet<String>();
HashMap<String, Integer> countMap = new HashMap<String, Integer>();
int count;
String key;
for (int i = 0; i < string.length(); i++) {
key = new String(string.charAt(i) + "");
charMap.add(key);
if(countMap.containsKey(key)) {
count = countMap.get(key);
countMap.put(key, count + 1);
}
else {
countMap.put(key, 1);
}
}
Iterator<String> iterator = charMap.iterator();
String resultStr = "";
while (iterator.hasNext()) {
key = iterator.next();
count = countMap.get(key);
if(count > 1) {
resultStr = resultStr + count + key;
}
else{
resultStr = resultStr + key;
}
}
System.out.println(resultStr);
}