Develop Reference

Reverse words without changing capitals or punctuation

Create a program with the lowest amount of characters to reverse each word in a string while keeping the order of the words, as well as punctuation and capital letters, in their initial place.
By "Order of the words", I mean that each word is split by an empty space (" "), so contractions and such will be treated as one word. The apostrophe in contractions should stay in the same place. ("Don't" => "Tno'd").
(Punctuation means any characters that are not a-z, A-Z or whitespace*).
Numbers were removed from this list due to the fact that you cannot have capital numbers. Numbers are now treated as punctuation.
For example, for the input:
Hello, I am a fish.
it should output:
Olleh, I ma a hsif.
Notice that O, which is the first letter in the first word, is now capital, since H was capital before in the same location.
The comma and the period are also in the same place.
More examples:
This; Is Some Text!
would output
Siht; Si Emos Txet!
I've tried this:
public static String reverseWord(String input)
{
String words[]=input.split(" ");
StringBuilder result=new StringBuilder();
for (String string : words) {
String revStr = new StringBuilder(string).reverse().toString();
result.append(revStr).append(" ");
}
return result.toString().trim();
}

I have tried to solve your problem. It's working fine for the examples I have checked :) Please look and let me know :)
public static void main(String[] args) {
System.out.println(reverseWord("This; Is Some Text!"));
}
public static boolean isAlphaNumeric(String s) {
return s != null && s.matches("^[a-zA-Z0-9]*$");
}
public static String reverseWord(String input)
{
String words[]=input.split(" ");
StringBuilder result=new StringBuilder();
int startIndex = 0;
int endIndex = 0;
for(int i = 0 ; i < input.length(); i++) {
if (isAlphaNumeric(Character.toString(input.charAt(i)))) {
endIndex++;
} else {
String string = input.substring(startIndex, endIndex);
startIndex = ++endIndex;
StringBuilder revStr = new StringBuilder("");
for (int j = 0; j < string.length(); j++) {
char charToAdd = string.charAt(string.length() - j - 1);
if (Character.isUpperCase(string.charAt(j))) {
revStr.append(Character.toUpperCase(charToAdd));
} else {
revStr.append(Character.toLowerCase(charToAdd));
}
}
result.append(revStr);
result.append(input.charAt(i));
}
}
if(endIndex>startIndex) // endIndex != startIndex
{
String string = input.substring(startIndex, endIndex);
result.append(string);
}
return result.toString().trim();
}
Call the reverseWord with your test string.
Hope it helps. Don't forget to mark it as right answer, if it is :)

Here is a proposal that follows your requirements. It may seem very long but its just comments and aerated code; and everybody loves comments.
public static String smartReverseWords(String input) {
StringBuilder finalString = new StringBuilder();
// Word accumulator, resetted after each "punctuation" (or anything different than a letter)
StringBuilder wordAcc = new StringBuilder();
int processedChars = 0;
for(char c : input.toCharArray()) {
// If not a whitespace nor the last character
if(!Character.isWhitespace(c)) {
// Accumulate letters
wordAcc.append(c);
// Have I reached the last character? Then finalize now:
if(processedChars == input.length()-1) {
reverseWordAndAppend(wordAcc, finalString);
}
}
else {
// Was a word accumulated?
if(wordAcc.length() > 0) {
reverseWordAndAppend(wordAcc, finalString);
}
// Append non-letter char to final string:
finalString.append(c);
}
processedChars++;
}
return finalString.toString();
}
private static void reverseWordAndAppend(StringBuilder wordAcc, StringBuilder finalString) {
// Then reverse it:
smartReverse(wordAcc); // a simple wordAcc.reverse() is not possible
// Append word to final string:
finalString.append(wordAcc.toString());
// Reset accumulator
wordAcc.setLength(0);
}
private static class Marker {
Integer position;
String character;
}
private static void smartReverse(StringBuilder wordAcc) {
char[] arr = wordAcc.toString().toCharArray();
wordAcc.setLength(0); // clean it for now
// Memorize positions of 'punctuation' + build array free of 'punctuation' in the same time:
List<Marker> mappedPosOfNonLetters = new ArrayList<>(); // order matters
List<Integer> mappedPosOfCapitals = new ArrayList<>(); // order matters
for (int i = 0; i < arr.length; i++) {
char c = arr[i];
if(!Character.isLetter(c)) {
Marker mark = new Marker();
mark.position = i;
mark.character = c+"";
mappedPosOfNonLetters.add(mark);
}
else {
if(Character.isUpperCase(c)) {
mappedPosOfCapitals.add(i);
}
wordAcc.append(Character.toLowerCase(c));
}
}
// Reverse cleansed word:
wordAcc.reverse();
// Reintroduce 'punctuation' at right place(s)
for (Marker mark : mappedPosOfNonLetters) {
wordAcc.insert(mark.position, mark.character);
}
// Restore capitals at right place(s)
for (Integer idx : mappedPosOfCapitals) {
wordAcc.setCharAt(idx,Character.toUpperCase(wordAcc.charAt(idx)));
}
}
EDIT
I've updated the code to take all your requirements into account. Indeed we have to make sure that "punctuation' stay in place (and capitals also) but also within a word, like a contraction.
Therefore given the following input string:
"Hello, I am on StackOverflow. Don't tell anyone."
The code produces this output:
"Olleh, I ma no WolfrEvokcats. Tno'd llet enoyna."

How to separate a String line with a paragraph to make text as a list

I have a really long text that looks like "123testes1233iambeginnerplshelp123 .." and I need to separate the line with the paragraph each time the program reads number.
So output should be like:
123tests
12333iambeninnerplshelp
123 ...

You can solve it using Regex. Everytime we are looking for patterns where number is followed by characters and if it is found, print it:
String text = "123testes1233stackoverflowwillsaveyou123dontworry";
String wordToFind = "\\d+[a-z]+";
Pattern word = Pattern.compile(wordToFind);
Matcher match = word.matcher(text);
while (match.find()) {
System.out.println(match.group());
}

One way to do it would be to use StringTokenizer. If you make the assumption that every output line must start with 123, even if the input doesn't start with it, it could be:
String input = "123testes1233iambeginnerplshelp123 ..";
String delimiter = "123";
StringTokenizer tokenizer = new StringTokenizer(input, delimiter);
while (tokenizer.hasMoreTokens()) {
String line = delimiter + tokenizer.nextToken();
System.out.println(line);
}

A simple approach (without any dependencies) would look something like this,
class Test {
public static void main (String[] args) throws java.lang.Exception
{
String a = "123testes1233iambeginnerplshelp123";
StringBuffer sb = new StringBuffer();
for (int i=0; i<a.length()-1; i++) {
while (i<a.length()-1 && !(!isNumber(a.charAt(i)) && isNumber(a.charAt(i+1)))) {
sb.append(a.substring(i,i+1));
i++;
}
sb.append(a.substring(i,i+1));
System.out.println(sb.toString());
sb.setLength(0);
}
}
private static boolean isNumber (char c) {
return ((int)c >=48) && ((int)c <= 57);
}
}

my solution
public StringSplitNum(){
String someString = "123testes1233iambeginnerplshelp123abc";
String regex = "((?<=[a-zA-Z])(?=[0-9]))|((?<=[0-9])(?=[a-zA-Z]))";
List arr = Arrays.asList(someString.split(regex));
for(int i=0; i< arr.size();i+=2){
System.out.println(arr.get(i)+ " " + arr.get(i+1));
}

Tokenize method: Split string into array

I've been really struggling with a programming assignment. Basically, we have to write a program that translates a sentence in English into one in Pig Latin. The first method we need is one to tokenize the string, and we are not allowed to use the Split method usually used in Java. I've been trying to do this for the past 2 days with no luck, here is what I have so far:
public class PigLatin
{
public static void main(String[] args)
{
String s = "Hello there my name is John";
Tokenize(s);
}
public static String[] Tokenize(String english)
{
String[] tokenized = new String[english.length()];
for (int i = 0; i < english.length(); i++)
{
int j= 0;
while (english.charAt(i) != ' ')
{
String m = "";
m = m + english.charAt(i);
if (english.charAt(i) == ' ')
{
j++;
}
else
{
break;
}
}
for (int l = 0; l < tokenized.length; l++) {
System.out.print(tokenized[l] + ", ");
}
}
return tokenized;
}
}
All this does is print an enormously long array of "null"s. If anyone can offer any input at all, I would reallllyyyy appreciate it!
Thank you in advance
Update: We are supposed to assume that there will be no punctuation or extra spaces, so basically whenever there is a space, it's a new word

If I understand your question, and what your Tokenize was intended to do; then I would start by writing a function to split the String
static String[] splitOnWhiteSpace(String str) {
List<String> al = new ArrayList<>();
StringBuilder sb = new StringBuilder();
for (char ch : str.toCharArray()) {
if (Character.isWhitespace(ch)) {
if (sb.length() > 0) {
al.add(sb.toString());
sb.setLength(0);
}
} else {
sb.append(ch);
}
}
if (sb.length() > 0) {
al.add(sb.toString());
}
String[] ret = new String[al.size()];
return al.toArray(ret);
}
and then print using Arrays.toString(Object[]) like
public static void main(String[] args) {
String s = "Hello there my name is John";
String[] words = splitOnWhiteSpace(s);
System.out.println(Arrays.toString(words));
}

If you're allowed to use the StringTokenizer Object (which I think is what the assignment is asking, it would look something like this:
StringTokenizer st = new StringTokenizer("this is a test");
while (st.hasMoreTokens()) {
System.out.println(st.nextToken());
}
which will produce the output:
this
is
a
test
Taken from here.
The string is split into tokens and stored in a stack. The while loop loops through the tokens, which is where you can apply the pig latin logic.

Some hints for you to do the "manual splitting" work.
There is a method String#indexOf(int ch, int fromIndex) to help you to find next occurrence of a character
There is a method String#substring(int beginIndex, int endIndex) to extract certain part of a string.
Here is some pseudo-code that show you how to split it (there are more safety handling that you need, I will leave that to you)
List<String> results = ...;
int startIndex = 0;
int endIndex = 0;
while (startIndex < inputString.length) {
endIndex = get next index of space after startIndex
if no space found {
endIndex = inputString.length
}
String result = get substring of inputString from startIndex to endIndex-1
results.add(result)
startIndex = endIndex + 1 // move startIndex to next position after space
}
// here, results contains all splitted words

String english = "hello my fellow friend"
ArrayList tokenized = new ArrayList<String>();
String m = "";
int j = 0; //index for tokenised array list.
for (int i = 0; i < english.length(); i++)
{
//the condition's position do matter here, if you
//change them, english.charAt(i) will give index
//out of bounds exception
while( i < english.length() && english.charAt(i) != ' ')
{
m = m + english.charAt(i);
i++;
}
//add to array list if there is some string
//if its only ' ', array will be empty so we are OK.
if(m.length() > 0 )
{
tokenized.add(m);
j++;
m = "";
}
}
//print the array list
for (int l = 0; l < tokenized.size(); l++) {
System.out.print(tokenized.get(l) + ", ");
}
This prints, "hello,my,fellow,friend,"
I used an array list since at the first sight the length of the array is not clear.

Reversing characters in each word in a sentence - Stack Implementation

This code is inside the main function:
Scanner input = new Scanner(System.in);
System.out.println("Type a sentence");
String sentence = input.next();
Stack<Character> stk = new Stack<Character>();
int i = 0;
while (i < sentence.length())
{
while (sentence.charAt(i) != ' ' && i < sentence.length() - 1)
{
stk.push(sentence.charAt(i));
i++;
}
stk.empty();
i++;
}
And this is the empty() function:
public void empty()
{
while (this.first != null)
System.out.print(this.pop());
}
It doesn't work properly, as by typing example sentence I am getting this output: lpmaxe. The first letter is missing and the loop stops instead of counting past the space to the next part of the sentence.
I am trying to achieve this:
This is a sentence ---> sihT si a ecnetnes

Per modifications to the original post, where the OP is now indicating that his goal is to reverse the letter order of the words within a sentence, but to leave the words in their initial positions.
The simplest way to do this, I think, is to make use of the String split function, iterate through the words, and reverse their orders.
String[] words = sentence.split(" "); // splits on the space between words
for (int i = 0; i < words.length; i++) {
String word = words[i];
System.out.print(reverseWord(word));
if (i < words.length-1) {
System.out.print(" "); // space after all words but the last
}
}
Where the method reverseWord is defined as:
public String reverseWord(String word) {
for( int i = 0; i < word.length(); i++) {
stk.push(word.charAt(i));
}
return stk.empty();
}
And where the empty method has been changed to:
public String empty() {
String stackWord = "";
while (this.first != null)
stackWord += this.pop();
return stackWord;
}
Original response
The original question indicated that the OP wanted to completely reverse the sentence.
You've got a double-looping construct where you don't really need it.
Consider this logic:
Read each character from the input string and push that character to the stack
When the input string is empty, pop each character from the stack and print it to screen.
So:
for( int i = 0; i < sentence.length(); i++) {
stk.push(sentence.charAt(i));
}
stk.empty();

I assume that what you want your code to do is to reverse each word in turn, not the entire string. So, given the input example sentence you want it to output elpmaxe ecnetnes not ecnetnes elpmaxe.
The reason that you see lpmaxe instead of elpmaxe is because your inner while-loop doesn't process the last character of the string since you have i < sentence.length() - 1 instead of i < sentence.length(). The reason that you only see a single word is because your sentence variable consists only of the first token of the input. This is what the method Scanner.next() does; it reads the next (by default) space-delimited token.
If you want to input a whole sentence, wrap up System.in as follows:
BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
and call reader.readLine().
Hope this helps.

Assuming you've already got your input in sentence and the Stack object is called stk, here's an idea:
char[] tokens = sentence.toCharArray();
for (char c : tokens) {
if (c == ' ') {
stk.empty();
System.out.print(c);
} else {
stk.add(c);
}
}
Thus, it will scan through one character at a time. If we hit a space character, we'll assume we've hit the end of a word, spit out that word in reverse, print that space character, then continue. Otherwise, we'll add the character to the stack and continue building the current word. (If you want to also allow punctuation like periods, commas, and the like, change if (c == ' ') { to something like if (c == ' ' || c == '.' || c == ',') { and so on.)
As for why you're only getting one word, darrenp already pointed it out. (Personally, I'd use a Scanner instead of a BufferedReader unless speed is an issue, but that's just my opinion.)

import java.util.StringTokenizer;
public class stringWork {
public static void main(String[] args) {
String s1 = "Hello World";
s1 = reverseSentence(s1);
System.out.println(s1);
s1 = reverseWord(s1);
System.out.println(s1);
}
private static String reverseSentence(String s1){
String s2 = "";
for(int i=s1.length()-1;i>=0;i--){
s2 += s1.charAt(i);
}
return s2;
}
private static String reverseWord(String s1){
String s2 = "";
StringTokenizer st = new StringTokenizer(s1);
while (st.hasMoreTokens()) {
s2 += reverseSentence(st.nextToken());
s2 += " ";
}
return s2;
}
}

public class ReverseofeachWordinaSentance {
/**
* #param args
*/
public static void main(String[] args) {
String source = "Welcome to the word reversing program";
for (String str : source.split(" ")) {
System.out.print(new StringBuilder(str).reverse().toString());
System.out.print(" ");
}
System.out.println("");
System.out.println("------------------------------------ ");
String original = "Welcome to the word reversing program";
wordReverse(original);
System.out.println("Orginal Sentence :::: "+original);
System.out.println("Reverse Sentence :::: "+wordReverse(original));
}
public static String wordReverse(String original){
StringTokenizer string = new StringTokenizer(original);
Stack<Character> charStack = new Stack<Character>();
while (string.hasMoreTokens()){
String temp = string.nextToken();
for (int i = 0; i < temp.length(); i ++){
charStack.push(temp.charAt(i));
}
charStack.push(' ');
}
StringBuilder result = new StringBuilder();
while(!charStack.empty()){
result.append(charStack.pop());
}
return result.toString();
}
}

public class reverseStr {
public static void main(String[] args) {
String testsa[] = { "", " ", " ", "a ", " a", " aa bd cs " };
for (String tests : testsa) {
System.out.println(tests + "|" + reverseWords2(tests) + "|");
}
}
public static String reverseWords2(String s) {
String[] sa;
String out = "";
sa = s.split(" ");
for (int i = 0; i < sa.length; i++) {
String word = sa[sa.length - 1 - i];
// exclude "" in splited array
if (!word.equals("")) {
//add space between two words
out += word + " ";
}
}
//exclude the last space and return when string is void
int n = out.length();
if (n > 0) {
return out.substring(0, out.length() - 1);
} else {
return "";
}
}
}
This can pass in leetcode

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");
}
}