Develop Reference

Java: Print a unique character in a string

I'm writing a program that will print the unique character in a string (entered through a scanner). I've created a method that tries to accomplish this but I keep getting characters that are not repeats, instead of a character (or characters) that is unique to the string. I want the unique letters only.
Here's my code:
import java.util.Scanner;
public class Sameness{
public static void main (String[]args){
Scanner kb = new Scanner (System.in);
String word = "";
System.out.println("Enter a word: ");
word = kb.nextLine();
uniqueCharacters(word);
}
public static void uniqueCharacters(String test){
String temp = "";
for (int i = 0; i < test.length(); i++){
if (temp.indexOf(test.charAt(i)) == - 1){
temp = temp + test.charAt(i);
}
}
System.out.println(temp + " ");
}
}
And here's sample output with the above code:
Enter a word:
nreena
nrea
The expected output would be: ra

Based on your desired output, you have to replace a character that initially has been already added when it has a duplicated later, so:
public static void uniqueCharacters(String test){
String temp = "";
for (int i = 0; i < test.length(); i++){
char current = test.charAt(i);
if (temp.indexOf(current) < 0){
temp = temp + current;
} else {
temp = temp.replace(String.valueOf(current), "");
}
}
System.out.println(temp + " ");
}

How about applying the KISS principle:
public static void uniqueCharacters(String test) {
System.out.println(test.chars().distinct().mapToObj(c -> String.valueOf((char)c)).collect(Collectors.joining()));
}

The accepted answer will not pass all the test case for example
input -"aaabcdd"
desired output-"bc"
but the accepted answer will give -abc
because the character a present odd number of times.
Here I have used ConcurrentHasMap to store character and the number of occurrences of character then removed the character if the occurrences is more than one time.
import java.util.concurrent.ConcurrentHashMap;
public class RemoveConductive {
public static void main(String[] args) {
String s="aabcddkkbghff";
String[] cvrtar=s.trim().split("");
ConcurrentHashMap<String,Integer> hm=new ConcurrentHashMap<>();
for(int i=0;i<cvrtar.length;i++){
if(!hm.containsKey(cvrtar[i])){
hm.put(cvrtar[i],1);
}
else{
hm.put(cvrtar[i],hm.get(cvrtar[i])+1);
}
}
for(String ele:hm.keySet()){
if(hm.get(ele)>1){
hm.remove(ele);
}
}
for(String key:hm.keySet()){
System.out.print(key);
}
}
}

Though to approach a solution I would suggest you to try and use a better data structure and not just string. Yet, you can simply modify your logic to delete already existing duplicates using an else as follows :
public static void uniqueCharacters(String test) {
String temp = "";
for (int i = 0; i < test.length(); i++) {
char ch = test.charAt(i);
if (temp.indexOf(ch) == -1) {
temp = temp + ch;
} else {
temp.replace(String.valueOf(ch),""); // added this to your existing code
}
}
System.out.println(temp + " ");
}

This is an interview question. Find Out all the unique characters of a string.
Here is the complete solution. The code itself is self explanatory.
public class Test12 {
public static void main(String[] args) {
String a = "ProtijayiGiniGina";
allunique(a);
}
private static void allunique(String a) {
int[] count = new int[256];// taking count of characters
for (int i = 0; i < a.length(); i++) {
char ch = a.charAt(i);
count[ch]++;
}
for (int i = 0; i < a.length(); i++) {
char chh = a.charAt(i);
// character which has arrived only one time in the string will be printed out
if (count[chh] == 1) {
System.out.println("index => " + i + " and unique character => " + a.charAt(i));
}
}
}// unique
}
In Python :
def firstUniqChar(a):
count = [0] *256
for i in a: count[ord(i)] += 1
element = ""
for item in a:
if (count[ord(item)] == 1):
element = item;
break;
return element
a = "GiniGinaProtijayi";
print(firstUniqChar(a)) # output is P

public static String input = "10 5 5 10 6 6 2 3 1 3 4 5 3";
public static void uniqueValue (String numbers) {
String [] str = input.split(" ");
Set <String> unique = new HashSet <String> (Arrays.asList(str));
System.out.println(unique);
for (String value:unique) {
int count = 0;
for ( int i= 0; i<str.length; i++) {
if (value.equals(str[i])) {
count++;
}
}
System.out.println(value+"\t"+count);
}
}
public static void main(String [] args) {
uniqueValue(input);
}

Step1: To find the unique characters in a string, I have first taken the string from user.
Step2: Converted the input string to charArray using built in function in java.
Step3: Considered two HashSet (set1 for storing all characters even if it is getting repeated, set2 for storing only unique characters.
Step4 : Run for loop over the array and check that if particular character is not there in set1 then add it to both set1 and set2. if that particular character is already there in set1 then add it to set1 again but remove it from set2.( This else part is useful when particular character is getting repeated odd number of times).
Step5 : Now set2 will have only unique characters. Hence, just print that set2.
public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
String str = input.next();
char arr[] = str.toCharArray();
HashSet<Character> set1=new HashSet<Character>();
HashSet<Character> set2=new HashSet<Character>();
for(char i:arr)
{
if(set1.contains(i))
{
set1.add(i);
set2.remove(i);
}
else
{
set1.add(i);
set2.add(i);
}
}
System.out.println(set2);
}

I would store all the characters of the string in an array that you will loop through to check if the current characters appears there more than once. If it doesn't, then add it to temp.
public static void uniqueCharacters(String test) {
String temp = "";
char[] array = test.toCharArray();
int count; //keep track of how many times the character exists in the string
outerloop: for (int i = 0; i < test.length(); i++) {
count = 0; //reset the count for every new letter
for(int j = 0; j < array.length; j++) {
if(test.charAt(i) == array[j])
count++;
if(count == 2){
count = 0;
continue outerloop; //move on to the next letter in the string; this will skip the next two lines below
}
}
temp += test.charAt(i);
System.out.println("Adding.");
}
System.out.println(temp);
}
I have added comments for some more detail.

import java.util.*;
import java.lang.*;
class Demo
{
public static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
System.out.println("Enter String");
String s1=sc.nextLine();
try{
HashSet<Object> h=new HashSet<Object>();
for(int i=0;i<s1.length();i++)
{
h.add(s1.charAt(i));
}
Iterator<Object> itr=h.iterator();
while(itr.hasNext()){
System.out.println(itr.next());
}
}
catch(Exception e)
{
System.out.println("error");
}
}
}

If you don't want to use additional space:
String abc="developer";
System.out.println("The unique characters are-");
for(int i=0;i<abc.length();i++)
{
for(int j=i+1;j<abc.length();j++)
{
if(abc.charAt(i)==abc.charAt(j))
abc=abc.replace(String.valueOf(abc.charAt(j))," ");
}
}
System.out.println(abc);
Time complexity O(n^2) and no space.

This String algorithm is used to print unique characters in a string.It runs in O(n) runtime where n is the length of the string.It supports ASCII characters only.
static String printUniqChar(String s) {
StringBuilder buildUniq = new StringBuilder();
boolean[] uniqCheck = new boolean[128];
for (int i = 0; i < s.length(); i++) {
if (!uniqCheck[s.charAt(i)]) {
uniqCheck[s.charAt(i)] = true;
if (uniqCheck[s.charAt(i)])
buildUniq.append(s.charAt(i));
}
}

public class UniqueCharactersInString {
public static void main(String []args){
String input = "aabbcc";
String output = uniqueString(input);
System.out.println(output);
}
public static String uniqueString(String s){
HashSet<Character> uniques = new HashSet<>();
uniques.add(s.charAt(0));
String out = "";
out += s.charAt(0);
for(int i =1; i < s.length(); i++){
if(!uniques.contains(s.charAt(i))){
uniques.add(s.charAt(i));
out += s.charAt(i);
}
}
return out;
}
}
What would be the inneficiencies of this answer? How does it compare to other answers?

Based on your desired output you can replace each character already present with a blank character.
public static void uniqueCharacters(String test){
String temp = "";
for(int i = 0; i < test.length(); i++){
if (temp.indexOf(test.charAt(i)) == - 1){
temp = temp + test.charAt(i);
} else {
temp.replace(String.valueOf(temp.charAt(i)), "");
}
}
System.out.println(temp + " ");
}

public void uniq(String inputString) {
String result = "";
int inputStringLen = inputStr.length();
int[] repeatedCharacters = new int[inputStringLen];
char inputTmpChar;
char tmpChar;
for (int i = 0; i < inputStringLen; i++) {
inputTmpChar = inputStr.charAt(i);
for (int j = 0; j < inputStringLen; j++) {
tmpChar = inputStr.charAt(j);
if (inputTmpChar == tmpChar)
repeatedCharacters[i]++;
}
}
for (int k = 0; k < inputStringLen; k++) {
inputTmpChar = inputStr.charAt(k);
if (repeatedCharacters[k] == 1)
result = result + inputTmpChar + " ";
}
System.out.println ("Unique characters: " + result);
}
In first for loop I count the number of times the character repeats in the string. In the second line I am looking for characters repetitive once.

how about this :)
for (int i=0; i< input.length();i++)
if(input.indexOf(input.charAt(i)) == input.lastIndexOf(input.charAt(i)))
System.out.println(input.charAt(i) + " is unique");

package extra;
public class TempClass {
public static void main(String[] args) {
// TODO Auto-generated method stub
String abcString="hsfj'pwue2hsu38bf74sa';fwe'rwe34hrfafnosdfoasq7433qweid";
char[] myCharArray=abcString.toCharArray();
TempClass mClass=new TempClass();
mClass.countUnique(myCharArray);
mClass.countEach(myCharArray);
}
/**
* This is the program to find unique characters in array.
* #add This is nice.
* */
public void countUnique(char[] myCharArray) {
int arrayLength=myCharArray.length;
System.out.println("Array Length is: "+arrayLength);
char[] uniqueValues=new char[myCharArray.length];
int uniqueValueIndex=0;
int count=0;
for(int i=0;i<arrayLength;i++) {
for(int j=0;j<arrayLength;j++) {
if (myCharArray[i]==myCharArray[j] && i!=j) {
count=count+1;
}
}
if (count==0) {
uniqueValues[uniqueValueIndex]=myCharArray[i];
uniqueValueIndex=uniqueValueIndex+1;
count=0;
}
count=0;
}
for(char a:uniqueValues) {
System.out.println(a);
}
}
/**
* This is the program to find count each characters in array.
* #add This is nice.
* */
public void countEach(char[] myCharArray) {
}
}

Here str will be your string to find the unique characters.
function getUniqueChars(str){
let uniqueChars = '';
for(let i = 0; i< str.length; i++){
for(let j= 0; j< str.length; j++) {
if(str.indexOf(str[i]) === str.lastIndexOf(str[j])) {
uniqueChars += str[i];
}
}
}
return uniqueChars;
}

public static void main(String[] args) {
String s = "aaabcdd";
char a[] = s.toCharArray();
List duplicates = new ArrayList();
List uniqueElements = new ArrayList();
for (int i = 0; i < a.length; i++) {
uniqueElements.add(a[i]);
for (int j = i + 1; j < a.length; j++) {
if (a[i] == a[j]) {
duplicates.add(a[i]);
break;
}
}
}
uniqueElements.removeAll(duplicates);
System.out.println(uniqueElements);
System.out.println("First Unique : "+uniqueElements.get(0));
}
Output :
[b, c]
First Unique : b

import java.util.*;
public class Sameness{
public static void main (String[]args){
Scanner kb = new Scanner (System.in);
String word = "";
System.out.println("Enter a word: ");
word = kb.nextLine();
uniqueCharacters(word);
}
public static void uniqueCharacters(String test){
for(int i=0;i<test.length();i++){
if(test.lastIndexOf(test.charAt(i))!=i)
test=test.replaceAll(String.valueOf(test.charAt(i)),"");
}
System.out.println(test);
}
}

public class Program02
{
public static void main(String[] args)
{
String inputString = "abhilasha";
for (int i = 0; i < inputString.length(); i++)
{
for (int j = i + 1; j < inputString.length(); j++)
{
if(inputString.toCharArray()[i] == inputString.toCharArray()[j])
{
inputString = inputString.replace(String.valueOf(inputString.charAt(j)), "");
}
}
}
System.out.println(inputString);
}
}

Counting words from array in a string

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

Generating String Tokens in Java

I want to generate possible tokens using forward traversal in Java. For example if I have a string "This is my car". I need to generate tokens
"This is my car"
"This is my"
"This is"
"This"
"is my car"
"is my"
"is"
"my car"
"my"
"car"
What is the best way to do this? Any examples? Thanks.

Here is another solution with split and nested loops:
public static void main(String[] args) {
String original = "this is my car";
String[] singleWords = original.split(" "); // split the String to get the single words
ArrayList<String> results = new ArrayList<String>(); // a container for all the possible sentences
for (int startWord = 0; startWord < singleWords.length; startWord++) { // starWords start with 0 and increment just until they reach the last word
for (int lastWord = singleWords.length; lastWord > startWord; lastWord--) { // last words start at the end and decrement just until they reached the first word
String next = "";
for (int i = startWord; i != lastWord; i++) { // put all words in one String (starting with the startWord and ending with the lastWord)
next += singleWords[i] + " ";
}
results.add(next); // add the next result to your result list
}
}
// this is just to check the results. All your sentences are now stored in the ArrayList results
for (String string : results) {
System.out.println("" + string);
}
}
and this was my result when I tested the method:
this is my car
this is my
this is
this
is my car
is my
is
my car
my
car

Use Guava:
String yourOriginalString = "This is my car";
final Set<String> originalWords =
Sets.newLinkedHashSet(
Splitter.on(CharMatcher.WHITESPACE).trimResults().split(yourOriginalString));
final Set<Set<String>> variations = Sets.powerSet(originalWords);
for (Set<String> variation : variations) {
System.out.println(Joiner.on(' ').join(variation));
}
Output:
This
is
This is
my
This my
is my
This is my
car
This car
is car
This is car
my car
This my car
is my car
This is my car

Here is a possible way:
//Just a method that seperates your String into an array of words based on the spaces
//I'll leave that for you to figure out how to make
String[] array = getSeperatedWords(<yourword>);
List<StringBuffer> bufferArray = new ArrayList<StringBuffer>();
for(int i = 0; i < array.length; i++){
StringBuffer nowWord = array[i];
for(int j = i; j < array.length; j++{
nowWord.append(array[j]);
}
bufferArray.add(nowWord);
}
for(int i = 0; i < bufferArray.length; i++){
System.out.print(bufferArray.get(i));
}

import java.util.Arrays;
public class Test {
public static void main(String[] args) {
String var = "This is my car";
permute(var);
}
public static void permute(String var) {
if(var.isEmpty())
return;
String[] arr = var.split(" ");
while(arr.length > 0) {
for(String str : arr) {
System.out.print(str + " ");
}
arr = (String[]) Arrays.copyOfRange(arr, 0, arr.length - 1);
System.out.println();
}
String[] original = var.split(" ");
permute(implodeArray((String[]) Arrays.copyOfRange(original, 1, original.length), " "));
}
public static String implodeArray(String[] inputArray, String glueString) {
String output = "";
if (inputArray.length > 0) {
StringBuilder sb = new StringBuilder();
sb.append(inputArray[0]);
for (int i=1; i<inputArray.length; i++) {
sb.append(glueString);
sb.append(inputArray[i]);
}
output = sb.toString();
}
return output;
}
}
Read this book, you will be a master on recursion: http://mitpress.mit.edu/sicp/

Java word count program

I am trying to make a program on word count which I have partially made and it is giving the correct result but the moment I enter space or more than one space in the string, the result of word count show wrong results because I am counting words on the basis of spaces used. I need help if there is a solution in a way that no matter how many spaces are I still get the correct result. I am mentioning the code below.
public class CountWords
{
public static void main (String[] args)
{
System.out.println("Simple Java Word Count Program");
String str1 = "Today is Holdiay Day";
int wordCount = 1;
for (int i = 0; i < str1.length(); i++)
{
if (str1.charAt(i) == ' ')
{
wordCount++;
}
}
System.out.println("Word count is = " + wordCount);
}
}

public static void main (String[] args) {
System.out.println("Simple Java Word Count Program");
String str1 = "Today is Holdiay Day";
String[] wordArray = str1.trim().split("\\s+");
int wordCount = wordArray.length;
System.out.println("Word count is = " + wordCount);
}
The ideas is to split the string into words on any whitespace character occurring any number of times.
The split function of the String class returns an array containing the words as its elements.
Printing the length of the array would yield the number of words in the string.

Two routes for this. One way would be to use regular expressions. You can find out more about regular expressions here. A good regular expression for this would be something like "\w+" Then count the number of matches.
If you don't want to go that route, you could have a boolean flag that remembers if the last character you've seen is a space. If it is, don't count it. So the center of the loop looks like this:
boolean prevCharWasSpace=true;
for (int i = 0; i < str1.length(); i++)
{
if (str1.charAt(i) == ' ') {
prevCharWasSpace=true;
}
else{
if(prevCharWasSpace) wordChar++;
prevCharWasSpace = false;
}
}
Update
Using the split technique is exactly equivalent to what's happening here, but it doesn't really explain why it works. If we go back to our CS theory, we want to construct a Finite State Automa (FSA) that counts words. That FSA may appear as:
If you look at the code, it implements this FSA exactly. The prevCharWasSpace keeps track of which state we're in, and the str1.charAt('i') is decideds which edge (or arrow) is being followed. If you use the split method, a regular expression equivalent of this FSA is constructed internally, and is used to split the string into an array.

Java does have StringTokenizer API and can be used for this purpose as below.
String test = "This is a test app";
int countOfTokens = new StringTokenizer(test).countTokens();
System.out.println(countOfTokens);
OR
in a single line as below
System.out.println(new StringTokenizer("This is a test app").countTokens());
StringTokenizer supports multiple spaces in the input string, counting only the words trimming unnecessary spaces.
System.out.println(new StringTokenizer("This is a test app").countTokens());
Above line also prints 5

You can use String.split (read more here) instead of charAt, you will get good results.
If you want to use charAt for some reason then try trimming the string before you count the words that way you won't have the extra space and an extra word

My implementation, not using StringTokenizer:
Map<String, Long> getWordCounts(List<String> sentences, int maxLength) {
Map<String, Long> commonWordsInEventDescriptions = sentences
.parallelStream()
.map(sentence -> sentence.replace(".", ""))
.map(string -> string.split(" "))
.flatMap(Arrays::stream)
.map(s -> s.toLowerCase())
.filter(word -> word.length() >= 2 && word.length() <= maxLength)
.collect(groupingBy(Function.identity(), counting()));
}
Then, you could call it like this, as an example:
getWordCounts(list, 9).entrySet().stream()
.filter(pair -> pair.getValue() <= 3 && pair.getValue() >= 1)
.findFirst()
.orElseThrow(() ->
new RuntimeException("No matching word found.")).getKey();
Perhaps flipping the method to return Map<Long, String> might be better.

Use split(regex) method. The result is an array of strings that was splited by regex.
String s = "Today is Holdiay Day";
System.out.println("Word count is = " + s.split(" ").length);

You need to read the file line by line and reduce the multiple occurences of the whitespaces appearing in your line to a single occurence and then count for the words. Following is a sample:
public static void main(String... args) throws IOException {
FileInputStream fstream = new FileInputStream("c:\\test.txt");
DataInputStream in = new DataInputStream(fstream);
BufferedReader br = new BufferedReader(new InputStreamReader(in));
String strLine;
int wordcount = 0;
while ((strLine = br.readLine()) != null) {
strLine = strLine.replaceAll("[\t\b]", "");
strLine = strLine.replaceAll(" {2,}", " ");
if (!strLine.isEmpty()){
wordcount = wordcount + strLine.split(" ").length;
}
}
System.out.println(wordcount);
in.close();
}

public class wordCOunt
{
public static void main(String ar[])
{
System.out.println("Simple Java Word Count Program");
String str1 = "Today is Holdiay Day";
int wordCount = 1;
for (int i = 0; i < str1.length(); i++)
{
if (str1.charAt(i) == ' '&& str1.charAt(i+1)!=' ')
{
wordCount++;
}
}
System.out.println("Word count is = " +(str1.length()- wordCount));
}
}

public class wordCount
{
public static void main(String ar[]) throws Exception
{
System.out.println("Simple Java Word Count Program");
int wordCount = 1,count=1;
BufferedReader br = new BufferedReader(new FileReader("C:/file.txt"));
String str2 = "", str1 = "";
while ((str1 = br.readLine()) != null) {
str2 += str1;
}
for (int i = 0; i < str2.length(); i++)
{
if (str2.charAt(i) == ' ' && str2.charAt(i+1)!=' ')
{
wordCount++;
}
}
System.out.println("Word count is = " +(wordCount));
}
}

you should make your code more generic by considering other word separators as well.. such as "," ";" etc.
public class WordCounter{
public int count(String input){
int count =0;
boolean incrementCounter = false;
for (int i=0; i<input.length(); i++){
if (isValidWordCharacter(input.charAt(i))){
incrementCounter = true;
}else if (incrementCounter){
count++;
incrementCounter = false;
}
}
if (incrementCounter) count ++;//if string ends with a valid word
return count;
}
private boolean isValidWordCharacter(char c){
//any logic that will help you identify a valid character in a word
// you could also have a method which identifies word separators instead of this
return (c >= 'A' && c<='Z') || (c >= 'a' && c<='z');
}
}

import com.google.common.base.Optional;
import com.google.common.base.Splitter;
import com.google.common.collect.HashMultiset;
import com.google.common.collect.ImmutableSet;
import com.google.common.collect.Multiset;
String str="Simple Java Word Count count Count Program";
Iterable<String> words = Splitter.on(" ").trimResults().split(str);
//google word counter
Multiset<String> wordsMultiset = HashMultiset.create();
for (String string : words) {
wordsMultiset.add(string.toLowerCase());
}
Set<String> result = wordsMultiset.elementSet();
for (String string : result) {
System.out.println(string+" X "+wordsMultiset.count(string));
}

public static int CountWords(String str){
if(str.length() == 0)
return 0;
int count =0;
for(int i=0;i< str.length();i++){
if(str(i) == ' ')
continue;
if(i > 0 && str.charAt(i-1) == ' '){
count++;
}
else if(i==0 && str.charAt(i) != ' '){
count++;
}
}
return count;
}

public class CountWords
{
public static void main (String[] args)
{
System.out.println("Simple Java Word Count Program");
String str1 = "Today is Holdiay Day";
int wordCount = 1;
for (int i = 0; i < str1.length(); i++)
{
if (str1.charAt(i) == ' ' && str1.charAt(i+1)!=' ')
{
wordCount++;
}
}
System.out.println("Word count is = " + wordCount));
}
}
This gives the correct result because if space comes twice or more then it can't increase wordcount. Enjoy.

try this
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
public class wordcount {
public static void main(String[] args) {
String s = "India is my country. I love India";
List<String> qw = new ArrayList<String>();
Map<String, Integer> mmm = new HashMap<String, Integer>();
for (String sp : s.split(" ")) {
qw.add(sp);
}
for (String num : qw) {
mmm.put(num, Collections.frequency(qw, num));
}
System.out.println(mmm);
}
}

To count total words Or to count total words without repeat word count
public static void main(String[] args) {
// TODO Auto-generated method stub
String test = "I am trying to make make make";
Pattern p = Pattern.compile("\\w+");
Matcher m = p.matcher(test);
HashSet<String> hs = new HashSet<>();
int i=0;
while (m.find()) {
i++;
hs.add(m.group());
}
System.out.println("Total words Count==" + i);
System.out.println("Count without Repetation ==" + hs.size());
}
}
Output :
Total words Count==7
Count without Repeatation ==5

Not sure if there is a drawback, but this worked for me...
Scanner input = new Scanner(System.in);
String userInput = input.nextLine();
String trimmed = userInput.trim();
int count = 1;
for (int i = 0; i < trimmed.length(); i++) {
if ((trimmed.charAt(i) == ' ') && (trimmed.charAt(i-1) != ' ')) {
count++;
}
}

You can use this code.It may help you:
public static void main (String[] args)
{
System.out.println("Simple Java Word Count Program");
String str1 = "Today is Holdiay Day";
int count=0;
String[] wCount=str1.split(" ");
for(int i=0;i<wCount.length;i++){
if(!wCount[i].isEmpty())
{
count++;
}
}
System.out.println(count);
}

String data = "This world is mine";
System.out.print(data.split("\\s+").length);

This could be as simple as using split and count variable.
public class SplitString {
public static void main(String[] args) {
int count=0;
String s1="Hi i love to code";
for(String s:s1.split(" "))
{
count++;
}
System.out.println(count);
}
}

public class TotalWordsInSentence {
public static void main(String[] args) {
String str = "This is sample sentence";
int NoOfWOrds = 1;
for (int i = 0; i<str.length();i++){
if ((str.charAt(i) == ' ') && (i!=0) && (str.charAt(i-1) != ' ')){
NoOfWOrds++;
}
}
System.out.println("Number of Words in Sentence: " + NoOfWOrds);
}
}
In this code, There wont be any problem regarding white-space in it.
just the simple for loop. Hope this helps...

To count specified words only like John, John99, John_John and John's only. Change regex according to yourself and count the specified words only.
public static int wordCount(String content) {
int count = 0;
String regex = "([a-zA-Z_’][0-9]*)+[\\s]*";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(content);
while(matcher.find()) {
count++;
System.out.println(matcher.group().trim()); //If want to display the matched words
}
return count;
}

class HelloWorld {
public static void main(String[] args) {
String str = "User is in for an interview";
int counter=0;
String arrStr[] = str.split(" ");
for (int i = 0; i< arrStr.length; i++){
String charStr = arrStr[i];
for(int j=0; j<charStr.length(); j++) {
if(charStr.charAt(j) =='i') {
counter++;
}
}
}
System.out.println("i " + counter);
}
}

public class CountWords {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.println("Enter the string :");
String str = sc.nextLine();
System.out.println("length is string is :"+str.length());
int worldCount = 1;
for(int i=0; i<str.length(); i++){
if(str.charAt(i) == ' '){
worldCount++;
}
}
System.out.println(worldCount);
}
}

The full program working is:
public class main {
public static void main(String[] args) {
logicCounter counter1 = new logicCounter();
counter1.counter("I am trying to make a program on word count which I have partially made and it is giving the correct result but the moment I enter space or more than one space in the string, the result of word count show wrong results because I am counting words on the basis of spaces used. I need help if there is a solution in a way that no matter how many spaces are I still get the correct result. I am mentioning the code below.");
}
}
public class logicCounter {
public void counter (String str) {
String str1 = str;
boolean space= true;
int i;
for ( i = 0; i < str1.length(); i++) {
if (str1.charAt(i) == ' ') {
space=true;
} else {
i++;
}
}
System.out.println("there are " + i + " letters");
}
}

I want to split string without using split function?

I want to split string without using split . can anybody solve my problem I am tried but
I cannot find the exact logic.

Since this seems to be a task designed as coding practice, I'll only guide. No code for you, sir, though the logic and the code aren't that far separated.
You will need to loop through each character of the string, and determine whether or not the character is the delimiter (comma or semicolon, for instance). If not, add it to the last element of the array you plan to return. If it is the delimiter, create a new empty string as the array's last element to start feeding your characters into.

I'm going to assume that this is homework, so I will only give snippets as hints:
Finding indices of all occurrences of a given substring
Here's an example of using indexOf with the fromIndex parameter to find all occurrences of a substring within a larger string:
String text = "012ab567ab0123ab";
// finding all occurrences forward: Method #1
for (int i = text.indexOf("ab"); i != -1; i = text.indexOf("ab", i+1)) {
System.out.println(i);
} // prints "3", "8", "14"
// finding all occurrences forward: Method #2
for (int i = -1; (i = text.indexOf("ab", i+1)) != -1; ) {
System.out.println(i);
} // prints "3", "8", "14"
String API links
int indexOf(String, int fromIndex)
Returns the index within this string of the first occurrence of the specified substring, starting at the specified index. If no such occurrence exists, -1 is returned.
Related questions
Searching for one string in another string
Extracting substrings at given indices out of a string
This snippet extracts substring at given indices out of a string and puts them into a List<String>:
String text = "0123456789abcdefghij";
List<String> parts = new ArrayList<String>();
parts.add(text.substring(0, 5));
parts.add(text.substring(3, 7));
parts.add(text.substring(9, 13));
parts.add(text.substring(18, 20));
System.out.println(parts); // prints "[01234, 3456, 9abc, ij]"
String[] partsArray = parts.toArray(new String[0]);
Some key ideas:
Effective Java 2nd Edition, Item 25: Prefer lists to arrays
Works especially nicely if you don't know how many parts there'll be in advance
String API links
String substring(int beginIndex, int endIndex)
Returns a new string that is a substring of this string. The substring begins at the specified beginIndex and extends to the character at index endIndex - 1.
Related questions
Fill array with List data

You do now that most of the java standard libraries are open source
In this case you can start here

Use String tokenizer to split strings in Java without split:
import java.util.StringTokenizer;
public class tt {
public static void main(String a[]){
String s = "012ab567ab0123ab";
String delims = "ab ";
StringTokenizer st = new StringTokenizer(s, delims);
System.out.println("No of Token = " + st.countTokens());
while (st.hasMoreTokens()) {
System.out.println(st.nextToken());
}
}
}

This is the right answer
import java.util.StringTokenizer;
public class tt {
public static void main(String a[]){
String s = "012ab567ab0123ab";
String delims = "ab ";
StringTokenizer st = new StringTokenizer(s, delims);
System.out.println("No of Token = " + st.countTokens());
while (st.hasMoreTokens())
{
System.out.println(st.nextToken());
}
}
}

/**
* My method split without javas split.
* Return array with words after mySplit from two texts;
* Uses trim.
*/
public class NoJavaSplit {
public static void main(String[] args) {
String text1 = "Some text for example ";
String text2 = " Second sentences ";
System.out.println(Arrays.toString(mySplit(text1, text2)));
}
private static String [] mySplit(String text1, String text2) {
text1 = text1.trim() + " " + text2.trim() + " ";
char n = ' ';
int massValue = 0;
for (int i = 0; i < text1.length(); i++) {
if (text1.charAt(i) == n) {
massValue++;
}
}
String[] splitArray = new String[massValue];
for (int i = 0; i < splitArray.length; ) {
for (int j = 0; j < text1.length(); j++) {
if (text1.charAt(j) == n) {
splitArray[i] = text1.substring(0, j);
text1 = text1.substring(j + 1, text1.length());
j = 0;
i++;
}
}
return splitArray;
}
return null;
}
}

you can try, the way i did `{
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String str = sc.nextLine();
for(int i = 0; i <str.length();i++) {
if(str.charAt(i)==' ') { // whenever it found space it'll create separate words from string
System.out.println();
continue;
}
System.out.print(str.charAt(i));
}
sc.close();
}`

The logic is: go through the whole string starting from first character and whenever you find a space copy the last part to a new string.. not that hard?

The way to go is to define the function you need first. In this case, it would probably be:
String[] split(String s, String separator)
The return type doesn't have to be an array. It can also be a list:
List<String> split(String s, String separator)
The code would then be roughly as follows:
start at the beginning
find the next occurence of the delimiter
the substring between the end of the previous delimiter and the start of the current delimiter is added to the result
continue with step 2 until you have reached the end of the string
There are many fine points that you need to consider:
What happens if the string starts or ends with the delimiter?
What if multiple delimiters appear next to each other?
What should be the result of splitting the empty string? (1 empty field or 0 fields)

You can do it using Java standard libraries.
Say the delimiter is : and
String s = "Harry:Potter"
int a = s.find(delimiter);
and then add
s.substring(start, a)
to a new String array.
Keep doing this till your start < string length
Should be enough I guess.

public class MySplit {
public static String[] mySplit(String text,String delemeter){
java.util.List<String> parts = new java.util.ArrayList<String>();
text+=delemeter;
for (int i = text.indexOf(delemeter), j=0; i != -1;) {
parts.add(text.substring(j,i));
j=i+delemeter.length();
i = text.indexOf(delemeter,j);
}
return parts.toArray(new String[0]);
}
public static void main(String[] args) {
String str="012ab567ab0123ab";
String delemeter="ab";
String result[]=mySplit(str,delemeter);
for(String s:result)
System.out.println(s);
}
}

public class WithoutSpit_method {
public static void main(String arg[])
{
char[]str;
String s="Computer_software_developer_gautam";
String s1[];
for(int i=0;i<s.length()-1;)
{
int lengh=s.indexOf("_",i);
if(lengh==-1)
{
lengh=s.length();
}
System.out.print(" "+s.substring(i,lengh));
i=lengh+1;
}
}
}
Result: Computer software developer gautam

Here is my way of doing with Scanner;
import java.util.Scanner;
public class spilt {
public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
System.out.print("Enter the String to be Spilted : ");
String st = input.nextLine();
Scanner str = new Scanner(st);
while (str.hasNext())
{
System.out.println(str.next());
}
}
}
Hope it Helps!!!!!

public class StringWitoutPre {
public static void main(String[] args) {
String str = "md taufique reja";
int len = str.length();
char ch[] = str.toCharArray();
String tmp = " ";
boolean flag = false;
for (int i = 0; i < str.length(); i++) {
if (ch[i] != ' ') {
tmp = tmp + ch[i];
flag = false;
} else {
flag = true;
}
if (flag || i == len - 1) {
System.out.println(tmp);
tmp = " ";
}
}
}
}

In Java8 we can use Pattern and get the things done in more easy way. Here is the code.
package com.company;
import java.util.regex.Pattern;
public class umeshtest {
public static void main(String a[]) {
String ss = "I'm Testing and testing the new feature";
Pattern.compile(" ").splitAsStream(ss).forEach(s -> System.out.println(s));
}
}

static void splitString(String s, int index) {
char[] firstPart = new char[index];
char[] secondPart = new char[s.length() - index];
int j = 0;
for (int i = 0; i < s.length(); i++) {
if (i < index) {
firstPart[i] = s.charAt(i);
} else {
secondPart[j] = s.charAt(i);
if (j < s.length()-index) {
j++;
}
}
}
System.out.println(firstPart);
System.out.println(secondPart);
}

import java.util.Scanner;
public class Split {
static Scanner in = new Scanner(System.in);
static void printArray(String[] array){
for (int i = 0; i < array.length; i++) {
if(i!=array.length-1)
System.out.print(array[i]+",");
else
System.out.println(array[i]);
}
}
static String delimeterTrim(String str){
char ch = str.charAt(str.length()-1);
if(ch=='.'||ch=='!'||ch==';'){
str = str.substring(0,str.length()-1);
}
return str;
}
private static String [] mySplit(String text, char reg, boolean delimiterTrim) {
if(delimiterTrim){
text = delimeterTrim(text);
}
text = text.trim() + " ";
int massValue = 0;
for (int i = 0; i < text.length(); i++) {
if (text.charAt(i) == reg) {
massValue++;
}
}
String[] splitArray = new String[massValue];
for (int i = 0; i < splitArray.length; ) {
for (int j = 0; j < text.length(); j++) {
if (text.charAt(j) == reg) {
splitArray[i] = text.substring(0, j);
text = text.substring(j + 1, text.length());
j = 0;
i++;
}
}
return splitArray;
}
return null;
}
public static void main(String[] args) {
System.out.println("Enter the sentence :");
String text = in.nextLine();
//System.out.println("Enter the regex character :");
//char regex = in.next().charAt(0);
System.out.println("Do you want to trim the delimeter ?");
String delch = in.next();
boolean ch = false;
if(delch.equalsIgnoreCase("yes")){
ch = true;
}
System.out.println("Output String array is : ");
printArray(mySplit(text,' ',ch));
}
}

Split a string without using split()
static String[] splitAString(String abc, char splitWith){
char[] ch=abc.toCharArray();
String temp="";
int j=0,length=0,size=0;
for(int i=0;i<abc.length();i++){
if(splitWith==abc.charAt(i)){
size++;
}
}
String[] arr=new String[size+1];
for(int i=0;i<ch.length;i++){
if(length>j){
j++;
temp="";
}
if(splitWith==ch[i]){
length++;
}else{
temp +=Character.toString(ch[i]);
}
arr[j]=temp;
}
return arr;
}
public static void main(String[] args) {
String[] arr=splitAString("abc-efg-ijk", '-');
for(int i=0;i<arr.length;i++){
System.out.println(arr[i]);
}
}
}

You cant split with out using split(). Your only other option is to get the strings char indexes and and get sub strings.