I want to split string without using split function?

I want to split string without using split function? - java

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.

Related

Long Form words to short form

How can I use indexOf() and CharAt() method to show the output like that?
"B R A M"
public class Test {
public static void main(String[] args) {
String s = new String("Business Requirement And Management");
String shortForm = "";
int index = 0;
int i=0;
while(i<s.length()) {
if(s.charAt(i) == ' '){
index = s.indexOf(" ");
shortForm += s.charAt(index+1);
}
i++;
}
System.out.println(shortForm);
}
}

I would use something like this:
StringBuilder sb = new StringBuilder();
String s = "Business Requirement And Management";
String[] splitted = s.split("\\s");
for(String str : splitted)
{
sb.append(str.charAt(0)).append(" ");
}
System.out.println(sb.toString());
Explanation:
We spilt the String at every whitespace character by calling split("\s") and store the new "substrings" in an array. Now iterate over the array and take the first character of each string and append it to the String.
Optionally you could use System.out.println(sb.toString().toUpperCase()) if you want to be certain that the initials are in upper case.

public class Test {
public static void main(String[] args) {
String s = new String("Business Requirement And Management");
String shortForm = "";
int index = 0;
while(s.length() > 0) {
index = s.indexOf(" ");
if(index == -1 || index == s.length() -1) break;
shortForm += s.charAt(index+1);
s = s.substring(index+1);
}
System.out.println(shortForm);
}
}

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);
}
}

Find the longest word in a String

Following is my code:
String LongestWord(String a)
{
int lw=0;
int use;
String lon="";
while (!(a.isEmpty()))
{
a=a.trim();
use=a.indexOf(" ");
if (use<0)
{
break;
}
String cut=a.substring(0,use);
if(cut.length()>lw)
{
lon=cut;
}
lw=lon.length();
a=a.replace(cut," ");
}
return lon;
}
The problem is that when I input a string like,
"a boy is playing in the park"
it returns the longest word as "ying" because when it replaces 'cut' with " " for the first time, it removes all the 'a'-s too, such that it becomes
" boy is pl ying in the p rk" after the first iteration of the loop
Please figure out what's wrong?
Thanks in advance!

You have already known the problem: the program does unwanted replacement.
Therefore, stop doing replacement.
In this program, the word examined is directly cut instead of using the harmful replacement.
String LongestWord(String a)
{
int lw=0;
int use;
String lon="";
while (!(a.isEmpty()))
{
a=a.trim();
use=a.indexOf(" ");
if (use<0)
{
break;
}
String cut=a.substring(0,use);
if(cut.length()>lw)
{
lon=cut;
}
lw=lon.length();
a=a.substring(use+1); // cut the word instead of doing harmful replacement
}
return lon;
}

You can use the split function to get an array of strings.
Than cycle that array to find the longest string and return it.
String LongestWord(String a) {
String[] parts = a.split(" ");
String longest = null;
for (String part : parts) {
if (longest == null || longest.length() < part.length()) {
longest = part;
}
}
return longest;
}

I would use arrays:
String[] parts = a.split(" ");
Then you can loop over parts, for each element (is a string) you can check length:
parts[i].length()
and find longest one.

I would use a Scanner to do this
String s = "the boy is playing in the parl";
int length = 0;
String word = "";
Scanner scan = new Scanner(s);
while(scan.hasNext()){
String temp = scan.next();
int tempLength = temp.length();
if(tempLength > length){
length = tempLength;
word = temp;
}
}
}
You check the length of each word, if it's longer then all the previous you store that word into the String "word"

Another way uses Streams.
Optional<String> max = Arrays.stream("a boy is playing in the park"
.split(" "))
.max((a, b) -> a.length() - b.length());
System.out.println("max = " + max);

if you are looking for not trivial Solution ,you can solve it without using split or map but with only one loop
static String longestWorld(String pharagragh) {
int maxLength = 0;
String word=null,longestWorld = null;
int startIndexOfWord = 0, endIndexOfWord;
int wordLength = 0;
for (int i = 0; i < pharagragh.length(); i++) {
if (pharagragh.charAt(i) == ' ') {
endIndexOfWord = i;
wordLength = endIndexOfWord - startIndexOfWord;
word = pharagragh.substring(startIndexOfWord, endIndexOfWord);
startIndexOfWord = endIndexOfWord + 1;
if (wordLength > maxLength) {
maxLength = wordLength;
longestWorld = word;
}
}
}
return longestWorld;
}
now lets test it
System.out.println(longestWorld("Hello Stack Overflow Welcome to Challenge World"));// output is Challenge

Try :
package testlongestword;
/**
*
* #author XOR
*/
public class TestLongestWord{
/**
* #param args the command line arguments
*/
public static void main(String[] args) {
System.out.println(LongestWord("a boy is playing in the park"));
}
public static String LongestWord(String str){
String[] words = str.split(" ");
int index = 0;
for(int i = 0; i < words.length; ++i){
final String current = words[i];
if(current.length() > words[index].length()){
index = i;
}
}
return words[index];
}
}

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");
}
}

How to upper case every first letter of word in a string? [duplicate]

This question already has answers here:
How to capitalize the first character of each word in a string
(51 answers)
Closed 3 years ago.
I have a string: "hello good old world" and i want to upper case every first letter of every word, not the whole string with .toUpperCase(). Is there an existing java helper which does the job?

Have a look at ACL WordUtils.
WordUtils.capitalize("your string") == "Your String"

Here is the code
String source = "hello good old world";
StringBuffer res = new StringBuffer();
String[] strArr = source.split(" ");
for (String str : strArr) {
char[] stringArray = str.trim().toCharArray();
stringArray[0] = Character.toUpperCase(stringArray[0]);
str = new String(stringArray);
res.append(str).append(" ");
}
System.out.print("Result: " + res.toString().trim());

sString = sString.toLowerCase();
sString = Character.toString(sString.charAt(0)).toUpperCase()+sString.substring(1);

i dont know if there is a function but this would do the job in case there is no exsiting one:
String s = "here are a bunch of words";
final StringBuilder result = new StringBuilder(s.length());
String[] words = s.split("\\s");
for(int i=0,l=words.length;i<l;++i) {
if(i>0) result.append(" ");
result.append(Character.toUpperCase(words[i].charAt(0)))
.append(words[i].substring(1));
}

import org.apache.commons.lang.WordUtils;
public class CapitalizeFirstLetterInString {
public static void main(String[] args) {
// only the first letter of each word is capitalized.
String wordStr = WordUtils.capitalize("this is first WORD capital test.");
//Capitalize method capitalizes only first character of a String
System.out.println("wordStr= " + wordStr);
wordStr = WordUtils.capitalizeFully("this is first WORD capital test.");
// This method capitalizes first character of a String and make rest of the characters lowercase
System.out.println("wordStr = " + wordStr );
}
}
Output :
This Is First WORD Capital Test.
This Is First Word Capital Test.

Here's a very simple, compact solution. str contains the variable of whatever you want to do the upper case on.
StringBuilder b = new StringBuilder(str);
int i = 0;
do {
b.replace(i, i + 1, b.substring(i,i + 1).toUpperCase());
i = b.indexOf(" ", i) + 1;
} while (i > 0 && i < b.length());
System.out.println(b.toString());
It's best to work with StringBuilder because String is immutable and it's inefficient to generate new strings for each word.

Trying to be more memory efficient than splitting the string into multiple strings, and using the strategy shown by Darshana Sri Lanka. Also, handles all white space between words, not just the " " character.
public static String UppercaseFirstLetters(String str)
{
boolean prevWasWhiteSp = true;
char[] chars = str.toCharArray();
for (int i = 0; i < chars.length; i++) {
if (Character.isLetter(chars[i])) {
if (prevWasWhiteSp) {
chars[i] = Character.toUpperCase(chars[i]);
}
prevWasWhiteSp = false;
} else {
prevWasWhiteSp = Character.isWhitespace(chars[i]);
}
}
return new String(chars);
}

String s = "java is an object oriented programming language.";
final StringBuilder result = new StringBuilder(s.length());
String words[] = s.split("\\ "); // space found then split it
for (int i = 0; i < words.length; i++)
{
if (i > 0){
result.append(" ");
}
result.append(Character.toUpperCase(words[i].charAt(0))).append(
words[i].substring(1));
}
System.out.println(result);
Output: Java Is An Object Oriented Programming Language.

Also you can take a look into StringUtils library. It has a bunch of cool stuff.

My code after reading a few above answers.
/**
* Returns the given underscored_word_group as a Human Readable Word Group.
* (Underscores are replaced by spaces and capitalized following words.)
*
* #param pWord
* String to be made more readable
* #return Human-readable string
*/
public static String humanize2(String pWord)
{
StringBuilder sb = new StringBuilder();
String[] words = pWord.replaceAll("_", " ").split("\\s");
for (int i = 0; i < words.length; i++)
{
if (i > 0)
sb.append(" ");
if (words[i].length() > 0)
{
sb.append(Character.toUpperCase(words[i].charAt(0)));
if (words[i].length() > 1)
{
sb.append(words[i].substring(1));
}
}
}
return sb.toString();
}

import java.util.Scanner;
public class CapitolizeOneString {
public static void main(String[] args)
{
Scanner scan = new Scanner(System.in);
System.out.print(" Please enter Your word = ");
String str=scan.nextLine();
printCapitalized( str );
} // end main()
static void printCapitalized( String str ) {
// Print a copy of str to standard output, with the
// first letter of each word in upper case.
char ch; // One of the characters in str.
char prevCh; // The character that comes before ch in the string.
int i; // A position in str, from 0 to str.length()-1.
prevCh = '.'; // Prime the loop with any non-letter character.
for ( i = 0; i < str.length(); i++ ) {
ch = str.charAt(i);
if ( Character.isLetter(ch) && ! Character.isLetter(prevCh) )
System.out.print( Character.toUpperCase(ch) );
else
System.out.print( ch );
prevCh = ch; // prevCh for next iteration is ch.
}
System.out.println();
}
} // end class

public class WordChangeInCapital{
public static void main(String[] args)
{
String s="this is string example";
System.out.println(s);
//this is input data.
//this example for a string where each word must be started in capital letter
StringBuffer sb=new StringBuffer(s);
int i=0;
do{
b.replace(i,i+1,sb.substring(i,i+1).toUpperCase());
i=b.indexOf(" ",i)+1;
} while(i>0 && i<sb.length());
System.out.println(sb.length());
}
}

package com.raj.samplestring;
/**
* #author gnagara
*/
public class SampleString {
/**
* #param args
*/
public static void main(String[] args) {
String[] stringArray;
String givenString = "ramu is Arr Good boy";
stringArray = givenString.split(" ");
for(int i=0; i<stringArray.length;i++){
if(!Character.isUpperCase(stringArray[i].charAt(0))){
Character c = stringArray[i].charAt(0);
Character change = Character.toUpperCase(c);
StringBuffer ss = new StringBuffer(stringArray[i]);
ss.insert(0, change);
ss.deleteCharAt(1);
stringArray[i]= ss.toString();
}
}
for(String e:stringArray){
System.out.println(e);
}
}
}

Here is an easy solution:
public class CapitalFirstLetters {
public static void main(String[] args) {
String word = "it's java, baby!";
String[] wordSplit;
String wordCapital = "";
wordSplit = word.split(" ");
for (int i = 0; i < wordSplit.length; i++) {
wordCapital = wordSplit[i].substring(0, 1).toUpperCase() + wordSplit[i].substring(1) + " ";
}
System.out.println(wordCapital);
}}

public String UpperCaseWords(String line)
{
line = line.trim().toLowerCase();
String data[] = line.split("\\s");
line = "";
for(int i =0;i< data.length;i++)
{
if(data[i].length()>1)
line = line + data[i].substring(0,1).toUpperCase()+data[i].substring(1)+" ";
else
line = line + data[i].toUpperCase();
}
return line.trim();
}

So much simpler with regexes:
Pattern spaces=Pattern.compile("\\s+[a-z]");
Matcher m=spaces.matcher(word);
StringBuilder capitalWordBuilder=new StringBuilder(word.substring(0,1).toUpperCase());
int prevStart=1;
while(m.find()) {
capitalWordBuilder.append(word.substring(prevStart,m.end()-1));
capitalWordBuilder.append(word.substring(m.end()-1,m.end()).toUpperCase());
prevStart=m.end();
}
capitalWordBuilder.append(word.substring(prevStart,word.length()));
Output for input: "this sentence Has Weird caps"
This Sentence Has Weird Caps

Develop Reference

Java is a programming language and computing platform first released by Sun Microsystems in 1995.

I want to split string without using split function? - java

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

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

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?

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.

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

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