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How to insert a char before each char in a string, but not before 'space' in Java?

This is what I got. It works for now, but if I type, for example, "I like bananas", I get 'pIp ppipkpep pbpapnpapnpaps', while I'm aiming to get 'pI pLpipkpe pbpapnpapnpaps.
Every solution I tried came down to using an 'if statement', trying to check if the character at said position in the original 'encText'is equal to ' ', and if so, making it equal to ' ' as well in the newText array before checking if the position required an 'p' or the char from the original array.. However, everytime I tried that, I'd get an array out of bounds exception.
static void pEncrypt() {
System.out.println("Adding P");
Scanner in = new Scanner(System.in);
String encText = in.nextLine();
int k = encText.length();
char[] charsEncText = encText.toCharArray();
char[] newText = new char[2*k];
int j = 1;
for (int i = 0; i < (k*2); i++) {
if (i%2 == 0) {
newText[i] = 'p';
} else {
newText[i] = charsEncText[i-j];
j++;
}
}
System.out.println(newText);
}

A simpler solution is to use replaceAll with a positive lookahead.
String str = "I like bananas";
String res = str.replaceAll("(?=[^ ])", "p");
System.out.println(res); // "pI plpipkpe pbpapnpapnpaps"
Demo

You can try this way.
static void pEncrypt() {
System.out.println("Adding P");
Scanner in = new Scanner(System.in);
String encText = in.nextLine();
int k = encText.length();
char[] charsEncText = encText.toCharArray();
char[] newText = new char[2*k];
int j=0;
for(int i=0;i<k;i++)
{
if(charsEncText[i]==' ')
{
newText[j]=charsEncText[i];
j++;
}
else{
newText[j]='p';
newText[j+1]=charsEncText[i];
j=j+2;
}
}
System.out.println(newText);
}

Assuming one does not need a char[], I would just concatenate to a String. Something like:
public static String pEncrypt(String org)
{
String ret = "";
for (int i = 0; i < org.length(); ++i) {
char ch = org.charAt(i);
if (ch != ' ') {
ret += "p" + ch;
}
else {
ret += ' ';
}
}
return ret;
}
Also, to make things a bit easier, it is generally a good idea to separate the I/O (i.e., the Scanner and the println) from the processing. That way, one can write test cases rather than attempting to keep inputting the information.
Sample Output:
helloworld ==> phpeplplpopwpoprplpd
I like bananas ==> pI plpipkpe pbpapnpapnpaps

Questions about easy "return" in JAVA

I want to make a code which has function of changing binary to decimal.
So i made a public long tonum()
and tried to return on Main method.
but there is anything shown on screen. Where is the problem?
Plz give me some hints.
public class Bitmap {
byte[] byteArr; //byte array for saving 0 or 1
char[] charArr; //char array for casting from string to byte array
public static void main(String[] args) throws Exception {
Bitmap test1 = new Bitmap("100110");
test1.tonum();
}
public Bitmap(String val) throws Exception {
byteArr = new byte[val.length()]; //To make the array length of Bitmap should e same as that of string
charArr = val.toCharArray(); //casting from string to char
for(int i = 0; i < charArr.length; i++) {
if (charArr[i] == '0')
byteArr[i] = 0;
else if (charArr[i] == '1')
byteArr[i] = 1;
else throw new Exception("Bitmap are should be sequences of zeros and ones!");
}
}
public long tonum() {
int temp = 0;
String str = "";
String str2 = "";
for (int i = 0; i < this.byteArr.length; i++){
temp = this.byteArr[i];
str = Integer.toString(temp);
str2 = str2 + str;
}
long decimal = (long)Integer.parseInt(str2,10);
System.out.println(str2);
return decimal;
}
}

long decimal = (long)Integer.parseInt(str2,10);
That doesn't change binary to decimal. It changes decimal to binary. And you don't have decimal in the first place, you have a string or presentation of binary. Try a radix of 2. But why you have both a char array and a byte array when all you do is deconstruct and reconstruct the original String is anybody's guess.

how to convert Lower case letters to upper case letters & and upper case letters to lower case letters

Alternately display any text that is typed in the textbox
// in either Capital or lowercase depending on the original
// letter changed. For example: CoMpUtEr will convert to
// cOmPuTeR and vice versa.
Switch.addActionListener(new ActionListener()
{
public void actionPerformed(ActionEvent e )
String characters = (SecondTextField.getText()); //String to read the user input
int length = characters.length(); //change the string characters to length
for(int i = 0; i < length; i++) //to check the characters of string..
{
char character = characters.charAt(i);
if(Character.isUpperCase(character))
{
SecondTextField.setText("" + characters.toLowerCase());
}
else if(Character.isLowerCase(character))
{
SecondTextField.setText("" + characters.toUpperCase()); //problem is here, how can i track the character which i already change above, means lowerCase**
}
}}
});

setText is changing the text content to exactly what you give it, not appending it.
Convert the String from the field first, then apply it directly...
String value = "This Is A Test";
StringBuilder sb = new StringBuilder(value);
for (int index = 0; index < sb.length(); index++) {
char c = sb.charAt(index);
if (Character.isLowerCase(c)) {
sb.setCharAt(index, Character.toUpperCase(c));
} else {
sb.setCharAt(index, Character.toLowerCase(c));
}
}
SecondTextField.setText(sb.toString());

You don't have to track whether you've already changed the character from upper to lower. Your code is already doing that since it's basically:
1 for each character x:
2 if x is uppercase:
3 convert x to lowercase
4 else:
5 if x is lowercase:
6 convert x to uppercase.
The fact that you have that else in there (on line 4) means that a character that was initially uppercase will never be checked in the second if statement (on line 5).
Example, start with A. Because that's uppercase, it will be converted to lowercase on line
3 and then you'll go back up to line 1 for the next character.
If you start with z, the if on line 2 will send you directly to line 5 where it will be converted to uppercase. Anything that's neither upper nor lowercase will fail both if statements and therefore remain untouched.

You can use StringUtils.swapCase() from org.apache.commons

This is a better method :-
void main()throws IOException
{
System.out.println("Enter sentence");
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String str = br.readLine();
String sentence = "";
for(int i=0;i<str.length();i++)
{
if(Character.isUpperCase(str.charAt(i))==true)
{
char ch2= (char)(str.charAt(i)+32);
sentence = sentence + ch2;
}
else if(Character.isLowerCase(str.charAt(i))==true)
{
char ch2= (char)(str.charAt(i)-32);
sentence = sentence + ch2;
}
else
sentence= sentence + str.charAt(i);
}
System.out.println(sentence);
}

The problem is that you are trying to set the value of SecondTextField after checking every single character in the original string. You should do the conversion "on the side", one character at a time, and only then set the result into the SecondTextField.
As you go through the original string, start composing the output from an empty string. Keep appending the character in the opposite case until you run out of characters. Once the output is ready, set it into SecondTextField.
You can make an output a String, set it to an empty string "", and append characters to it as you go. This will work, but that is an inefficient approach. A better approach would be using a StringBuilder class, which lets you change the string without throwing away the whole thing.

String name = "Vikash";
String upperCase = name.toUpperCase();
String lowerCase = name.toLowerCase();

This is a better approach without using any String function.
public static String ReverseCases(String str) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < str.length(); i++) {
char temp;
if (str.charAt(i) >= 'a' && str.charAt(i) <= 'z') {
temp = (char)(str.charAt(i) - 32);
}
else if (str.charAt(i) >= 'A' && str.charAt(i) <= 'Z'){
temp = (char)(str.charAt(i) + 32);
}
else {
temp = str.charAt(i);
}
sb.append(temp);
}
return sb.toString();
}

Here you are some other version:
public class Palindrom {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.println("Enter a word to check: ");
String checkWord = sc.nextLine();
System.out.println(isPalindrome(checkWord));
sc.close();
}
public static boolean isPalindrome(String str) {
StringBuilder secondSB = new StringBuilder();
StringBuilder sb = new StringBuilder();
sb.append(str);
for(int i = 0; i<sb.length();i++){
char c = sb.charAt(i);
if(Character.isUpperCase(c)){
sb.setCharAt(i, Character.toLowerCase(c));
}
}
secondSB.append(sb);
return sb.toString().equals(secondSB.reverse().toString());
}
}

StringBuilder b = new StringBuilder();
Scanner s = new Scanner(System.in);
String n = s.nextLine();
for(int i = 0; i < n.length(); i++) {
char c = n.charAt(i);
if(Character.isLowerCase(c) == true) {
b.append(String.valueOf(c).toUpperCase());
}
else {
b.append(String.valueOf(c).toLowerCase());
}
}
System.out.println(b);

Methods description:
*toLowerCase()* Returns a new string with all characters converted to lowercase.
*toUpperCase()* Returns a new string with all characters converted to uppercase.
For example:
"Welcome".toLowerCase() returns a new string, welcome
"Welcome".toUpperCase() returns a new string, WELCOME

If you look at characters a-z, you'll see that all of them have the 6th bit is set to 1. Where in A-Z 6th bit is not set.
A = 1000001 a = 1100001
B = 1000010 b = 1100010
C = 1000011 c = 1100011
D = 1000100 d = 1100100
...
Z = 1011010 z = 1111010
So all we need to do is to iterate through each character from a given string and then do XOR(^) with 32. In this way, the 6th bit can swap.
Look at the below code for simply changing the string case without using any if-else conditions.
public final class ChangeStringCase {
public static void main(String[] args) {
String str = "Hello World";
for (int i = 0; i < str.length(); i++) {
char ans = (char)(str.charAt(i) ^ 32);
System.out.print(ans); // Final Output: hELLO wORLD
}
}
}
Time Complexity: O(N) where N = Length of the string.
Space Complexity: O(1)

public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
String satr=scanner.nextLine();
String newString = "";
for (int i = 0; i < satr.length(); i++) {
if (Character.isUpperCase(satr.charAt(i))) {
newString+=Character.toLowerCase(satr.charAt(i));
}else newString += Character.toUpperCase(satr.charAt(i));
}
System.out.println(newString);
}

public class Toggle {
public static String toggle(String s) {
char[] ch = s.toCharArray();
for (int i = 0; i < s.length(); i++) {
char charat = ch[i];
if (Character.isUpperCase(charat)) {
charat = Character.toLowerCase(charat);
} else
charat = Character.toUpperCase(charat);
System.out.print(charat);
}
return s;
}
public static void main(String[] args) {
toggle("DivYa");
}
}

import java.util.Scanner;
class TestClass {
public static void main(String args[]) throws Exception {
Scanner s = new Scanner(System.in);
String str = s.nextLine();
char[] ch = str.toCharArray();
for (int i = 0; i < ch.length; i++) {
if (Character.isUpperCase(ch[i])) {
ch[i] = Character.toLowerCase(ch[i]);
} else {
ch[i] = Character.toUpperCase(ch[i]);
}
}
System.out.println(ch);
}
}

//This is to convert a letter from upper case to lower case
import java.util.Scanner;
public class ChangeCase {
public static void main(String[]args) {
String input;
Scanner sc= new Scanner(System.in);
System.out.println("Enter Letter from upper case");
input=sc.next();
String result;
result= input.toLowerCase();
System.out.println(result);
}
}

String str1,str2;
Scanner S=new Scanner(System.in);
str1=S.nextLine();
System.out.println(str1);
str2=S.nextLine();
str1=str1.concat(str2);
System.out.println(str1.toLowerCase());

Compare strings in java and remove the part of string where they are identical

I have two strings with me:
s1="MICROSOFT"
s2="APPLESOFT"
I need to compare the strings and remove the duplicate part (always towards the end) from the second string. So I should get "MICROSOFT" and "APPLE" as output.
I have compared both the strings character by character.
String s1 = "MICROSOFT";
String s2 = "APPLESOFT";
for(int j=0; j<s1.length(); j++)
{
char c1 = s1.charAt(j);
char c2 = s2.charAt(j);
if(c1==c2)
System.out.println("Match found!!!");
else
System.out.println("No match found!");
}
It should check the strings and if the two strings have same characters until the end of string, then I need to remove that redundant part, SOFT in this case, from the second string. But I can't think of how to proceed from here.
There can be more duplicates...but we have to remove only those which are continuously identical. if i have APPWWSOFT and APPLESOFT, i should get APPLE again in the second string since we got LE different than WW in between
Can you guys please help me out here?

Search and read about Longest Common Subsequence, you can find efficient algorithms to find out the LCS of two input strings. After finding the LCS of the input strings, it is easy to manipulate the inputs. For example, in your case an LCS algorithm will find "SOFT" as the LCS of these two strings, then you might check whether the LCS is in the final part of the 2nd input and then remove it easily. I hope this idea helps.
An example LCS code in Java is here, try it: http://introcs.cs.princeton.edu/java/96optimization/LCS.java.html
Example scenario (pseudocode):
input1: "MISROSOFT";
input2: "APPLESOFT";
execute LCS(input1, input2);
store the result in lcs, now lcs = "SOFT";
iterate over the characters of input2,
if a character exists in lcs then remove it from input2.

As far as I understand, you want to remove any identical characters from the two strings. By identical I mean: same position and same character(code). I think the following linear complexity solution is the simplest:
StringBuilder sb1 = new StringBuilder();
StringBuilder sb2 = new StringBuilder(); //if you want to remove the identical char
//only from one string you don't need the 2nd sb
char c;
for(int i = 0; i<Math.min(s1.length,s2.length);i++){
if((c = s1.charAt(i)) != s2.charAt(i)){
sb1.append(c);
}
}
return sb1.toString();

Try this algo- Create characters sequences of your first string and find it in second string.
performance -
Average case = (s1.length()-1)sq
public class SeqFind {
public static String searchReplace(String s1,String s2) {
String s3;
boolean brk=false;
for(int j=s1.length();j>0&&!brk;j--){
for (int i = j-4; i > 0; i--) {
String string = s1.substring( i,j);
if(s2.contains(string)){
System.out.println(s2+" - "+string+" "+s2.replace( string,""));
brk=true;
break;
}
}
}
return s3;
}
public static void main(String[] args) {
String s1 = "MICROSOFT";
String s2 = "APPLESOFT";
String s3 = searchReplace(s1,s2);
}
}
Out put -
APPLESOFT - SOFT - APPLE

public class Match {
public static void main(String[] args)
{
String s1="MICROSOFT";
String s2="APPLESOFT";
String[] s=new String[10];
String s3;
int j=0,k=0;
for(int i=s2.length();i>0;i--)
{
s[j]=s2.substring(k,s2.length());
if(s1.contains(s[j]))
{
s3=s2.substring(0,j);
System.out.println(s1+""+s3);
System.exit(0);
}
else
{
System.out.println("");
}
j++;
k++;
}
}
}
I have edited the code you can give it an another try.

try this, not tested thou
String s1 = "MICROSOFT";
String s2 = "APPLESOFT";
String s3="";
for(int j=0; j<s1.length(); j++)
{
if(s1.charAt(j)==s2.charAt(j)){
s3+=s1.charAt(j);
}
}
System.out.println(s1.replace(s3, " ") + " \n"+ s2.replace(s3, " "));

You should rather use StringBuffer if you want your String to be modified..
And in this case, you can have one extra StringBuffer, in which you can keep on appending non-matching character: -
StringBuffer s1 = new StringBuffer("MICROSOFT");
StringBuffer s2 = new StringBuffer("APPLESOFT");
StringBuffer s3 = new StringBuffer();
for(int j=0; j<s1.length(); j++)
{
char c1 = s1.charAt(j);
char c2 = s2.charAt(j);
if(c1==c2) {
System.out.println("Match found!!!");
} else {
System.out.println("No match found!");
s3.append(c1);
}
}
s1 = s3;
System.out.println(s1); // Prints "MICRO"

I have solved my problem after racking some brains off. Please feel free to correct/improve/refine my code. The code not only works for "MICROSOFT" and "APPLESOFT" inputs, but also for inputs like "APPWWSOFT" and "APPLESOFT" (i needed to remove the continuous duplicates from the end - SOFT in both the above inputs). I'm in the learning stage and I'll appreciate any valuable inputs.
public class test
{
public static void main(String[] args)
{
String s1 = "MICROSOFT";
String s2 = "APPLESOFT";
int counter1=0;
int counter2=0;
String[] test = new String[100];
test[0]="";
for(int j=0; j<s1.length(); j++)
{
char c1 = s1.charAt(j);
char c2 = s2.charAt(j);
if(c1==c2)
{
if(counter1==counter2)
{
//System.out.println("Match found!!!");
test[0]=test[0]+c2;
counter2++;
//System.out.println("Counter 2: "+counter2);
}
else
test[0]="";
}
else
{
//System.out.print("No match found!");
//System.out.println("Counter 2: "+counter2);
counter2=counter1+1;
test[0]="";
}
counter1++;
//System.out.println("Counter 1: "+counter1);
}
System.out.println(test[0]);
System.out.println(s2.replaceAll(test[0]," "));
}
}

Removing duplicates from a String in Java

I am trying to iterate through a string in order to remove the duplicates characters.
For example the String aabbccdef should become abcdef
and the String abcdabcd should become abcd
Here is what I have so far:
public class test {
public static void main(String[] args) {
String input = new String("abbc");
String output = new String();
for (int i = 0; i < input.length(); i++) {
for (int j = 0; j < output.length(); j++) {
if (input.charAt(i) != output.charAt(j)) {
output = output + input.charAt(i);
}
}
}
System.out.println(output);
}
}
What is the best way to do this?

Convert the string to an array of char, and store it in a LinkedHashSet. That will preserve your ordering, and remove duplicates. Something like:
String string = "aabbccdefatafaz";
char[] chars = string.toCharArray();
Set<Character> charSet = new LinkedHashSet<Character>();
for (char c : chars) {
charSet.add(c);
}
StringBuilder sb = new StringBuilder();
for (Character character : charSet) {
sb.append(character);
}
System.out.println(sb.toString());

Using Stream makes it easy.
noDuplicates = Arrays.asList(myString.split(""))
.stream()
.distinct()
.collect(Collectors.joining());
Here is some more documentation about Stream and all you can do with
it :
https://docs.oracle.com/javase/8/docs/api/java/util/stream/package-summary.html
The 'description' part is very instructive about the benefits of Streams.

Try this simple solution:
public String removeDuplicates(String input){
String result = "";
for (int i = 0; i < input.length(); i++) {
if(!result.contains(String.valueOf(input.charAt(i)))) {
result += String.valueOf(input.charAt(i));
}
}
return result;
}

I would use the help of LinkedHashSet. Removes dups (as we are using a Set, maintains the order as we are using linked list impl). This is kind of a dirty solution. there might be even a better way.
String s="aabbccdef";
Set<Character> set=new LinkedHashSet<Character>();
for(char c:s.toCharArray())
{
set.add(Character.valueOf(c));
}

Create a StringWriter. Run through the original string using charAt(i) in a for loop. Maintain a variable of char type keeping the last charAt value. If you iterate and the charAt value equals what is stored in that variable, don't add to the StringWriter. Finally, use the StringWriter.toString() method and get a string, and do what you need with it.

Here is an improvement to the answer by Dave.
It uses HashSet instead of the slightly more costly LinkedHashSet, and reuses the chars buffer for the result, eliminating the need for a StringBuilder.
String string = "aabbccdefatafaz";
char[] chars = string.toCharArray();
Set<Character> present = new HashSet<>();
int len = 0;
for (char c : chars)
if (present.add(c))
chars[len++] = c;
System.out.println(new String(chars, 0, len)); // abcdeftz

Java 8 has a new String.chars() method which returns a stream of characters in the String. You can use stream operations to filter out the duplicate characters like so:
String out = in.chars()
.mapToObj(c -> Character.valueOf((char) c)) // bit messy as chars() returns an IntStream, not a CharStream (which doesn't exist)
.distinct()
.map(Object::toString)
.collect(Collectors.joining(""));

String input = "AAAB";
String output = "";
for (int index = 0; index < input.length(); index++) {
if (input.charAt(index % input.length()) != input
.charAt((index + 1) % input.length())) {
output += input.charAt(index);
}
}
System.out.println(output);
but you cant use it if the input has the same elements, or if its empty!

Code to remove the duplicate characters in a string without using any additional buffer. NOTE: One or two additional variables are fine. An extra array is not:
import java.util.*;
public class Main{
public static char[] removeDupes(char[] arr){
if (arr == null || arr.length < 2)
return arr;
int len = arr.length;
int tail = 1;
for(int x = 1; x < len; x++){
int y;
for(y = 0; y < tail; y++){
if (arr[x] == arr[y]) break;
}
if (y == tail){
arr[tail] = arr[x];
tail++;
}
}
return Arrays.copyOfRange(arr, 0, tail);
}
public static char[] bigArr(int len){
char[] arr = new char[len];
Random r = new Random();
String alphabet = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890!##$%^&*()-=_+[]{}|;:',.<>/?`~";
for(int x = 0; x < len; x++){
arr[x] = alphabet.charAt(r.nextInt(alphabet.length()));
}
return arr;
}
public static void main(String args[]){
String result = new String(removeDupes(new char[]{'a', 'b', 'c', 'd', 'a'}));
assert "abcd".equals(result) : "abcda should return abcd but it returns: " + result;
result = new String(removeDupes(new char[]{'a', 'a', 'a', 'a'}));
assert "a".equals(result) : "aaaa should return a but it returns: " + result;
result = new String(removeDupes(new char[]{'a', 'b', 'c', 'a'}));
assert "abc".equals(result) : "abca should return abc but it returns: " + result;
result = new String(removeDupes(new char[]{'a', 'a', 'b', 'b'}));
assert "ab".equals(result) : "aabb should return ab but it returns: " + result;
result = new String(removeDupes(new char[]{'a'}));
assert "a".equals(result) : "a should return a but it returns: " + result;
result = new String(removeDupes(new char[]{'a', 'b', 'b', 'a'}));
assert "ab".equals(result) : "abba should return ab but it returns: " + result;
char[] arr = bigArr(5000000);
long startTime = System.nanoTime();
System.out.println("2: " + new String(removeDupes(arr)));
long endTime = System.nanoTime();
long duration = (endTime - startTime);
System.out.println("Program took: " + duration + " nanoseconds");
System.out.println("Program took: " + duration/1000000000 + " seconds");
}
}
How to read and talk about the above code:
The method called removeDupes takes an array of primitive char called arr.
arr is returned as an array of primitive characters "by value". The arr passed in is garbage collected at the end of Main's member method removeDupes.
The runtime complexity of this algorithm is O(n) or more specifically O(n+(small constant)) the constant being the unique characters in the entire array of primitive chars.
The copyOfRange does not increase runtime complexity significantly since it only copies a small constant number of items. The char array called arr is not stepped all the way through.
If you pass null into removeDupes, the method returns null.
If you pass an empty array of primitive chars or an array containing one value, that unmodified array is returned.
Method removeDupes goes about as fast as physically possible, fully utilizing the L1 and L2 cache, so Branch redirects are kept to a minimum.
A 2015 standard issue unburdened computer should be able to complete this method with an primitive char array containing 500 million characters between 15 and 25 seconds.
Explain how this code works:
The first part of the array passed in is used as the repository for the unique characters that are ultimately returned. At the beginning of the function the answer is: "the characters between 0 and 1" as between 0 and tail.
We define the variable y outside of the loop because we want to find the first location where the array index that we are looking at has been duplicated in our repository. When a duplicate is found, it breaks out and quits, the y==tail returns false and the repository is not contributed to.
when the index x that we are peeking at is not represented in our repository, then we pull that one and add it to the end of our repository at index tail and increment tail.
At the end, we return the array between the points 0 and tail, which should be smaller or equal to in length to the original array.
Talking points exercise for coder interviews:
Will the program behave differently if you change the y++ to ++y? Why or why not.
Does the array copy at the end represent another 'N' pass through the entire array making runtime complexity O(n*n) instead of O(n) ? Why or why not.
Can you replace the double equals comparing primitive characters with a .equals? Why or why not?
Can this method be changed in order to do the replacements "by reference" instead of as it is now, "by value"? Why or why not?
Can you increase the efficiency of this algorithm by sorting the repository of unique values at the beginning of 'arr'? Under which circumstances would it be more efficient?

public class RemoveRepeated4rmString {
public static void main(String[] args) {
String s = "harikrishna";
String s2 = "";
for (int i = 0; i < s.length(); i++) {
Boolean found = false;
for (int j = 0; j < s2.length(); j++) {
if (s.charAt(i) == s2.charAt(j)) {
found = true;
break; //don't need to iterate further
}
}
if (found == false) {
s2 = s2.concat(String.valueOf(s.charAt(i)));
}
}
System.out.println(s2);
}
}

public static void main(String a[]){
String name="Madan";
System.out.println(name);
StringBuilder sb=new StringBuilder(name);
for(int i=0;i<name.length();i++){
for(int j=i+1;j<name.length();j++){
if(name.charAt(i)==name.charAt(j)){
sb.deleteCharAt(j);
}
}
}
System.out.println("After deletion :"+sb+"");
}

import java.util.Scanner;
public class dublicate {
public static void main(String... a) {
System.out.print("Enter the String");
Scanner Sc = new Scanner(System.in);
String st=Sc.nextLine();
StringBuilder sb=new StringBuilder();
boolean [] bc=new boolean[256];
for(int i=0;i<st.length();i++)
{
int index=st.charAt(i);
if(bc[index]==false)
{
sb.append(st.charAt(i));
bc[index]=true;
}
}
System.out.print(sb.toString());
}
}

To me it looks like everyone is trying way too hard to accomplish this task. All we are concerned about is that it copies 1 copy of each letter if it repeats. Then because we are only concerned if those characters repeat one after the other the nested loops become arbitrary as you can just simply compare position n to position n + 1. Then because this only copies things down when they're different, to solve for the last character you can either append white space to the end of the original string, or just get it to copy the last character of the string to your result.
String removeDuplicate(String s){
String result = "";
for (int i = 0; i < s.length(); i++){
if (i + 1 < s.length() && s.charAt(i) != s.charAt(i+1)){
result = result + s.charAt(i);
}
if (i + 1 == s.length()){
result = result + s.charAt(i);
}
}
return result;
}

String str1[] ="Hi helloo helloo oooo this".split(" ");
Set<String> charSet = new LinkedHashSet<String>();
for (String c: str1)
{
charSet.add(c);
}
StringBuilder sb = new StringBuilder();
for (String character : charSet)
{
sb.append(character);
}
System.out.println(sb.toString());

I think working this way would be more easy,,,
Just pass a string to this function and the job is done :) .
private static void removeduplicate(String name)
{ char[] arr = name.toCharArray();
StringBuffer modified =new StringBuffer();
for(char a:arr)
{
if(!modified.contains(Character.toString(a)))
{
modified=modified.append(Character.toString(a)) ;
}
}
System.out.println(modified);
}

public class RemoveDuplicatesFromStingsMethod1UsingLoops {
public static void main(String[] args) {
String input = new String("aaabbbcccddd");
String output = "";
for (int i = 0; i < input.length(); i++) {
if (!output.contains(String.valueOf(input.charAt(i)))) {
output += String.valueOf(input.charAt(i));
}
}
System.out.println(output);
}
}
output: abcd

You can't. You can create a new String that has duplicates removed. Why aren't you using StringBuilder (or StringBuffer, presumably)?
You can run through the string and store the unique characters in a char[] array, keeping track of how many unique characters you've seen. Then you can create a new String using the String(char[], int, int) constructor.
Also, the problem is a little ambiguous—does “duplicates” mean adjacent repetitions? (In other words, what should happen with abcab?)

Oldschool way (as we wrote such a tasks in Apple ][ Basic, adapted to Java):
int i,j;
StringBuffer str=new StringBuffer();
Scanner in = new Scanner(System.in);
System.out.print("Enter string: ");
str.append(in.nextLine());
for (i=0;i<str.length()-1;i++){
for (j=i+1;j<str.length();j++){
if (str.charAt(i)==str.charAt(j))
str.deleteCharAt(j);
}
}
System.out.println("Removed non-unique symbols: " + str);

Here is another logic I'd like to share. You start comparing from midway of the string length and go backward.
Test with:
input = "azxxzy";
output = "ay";
String removeMidway(String input){
cnt = cnt+1;
StringBuilder str = new StringBuilder(input);
int midlen = str.length()/2;
for(int i=midlen-1;i>0;i--){
for(int j=midlen;j<str.length()-1;j++){
if(str.charAt(i)==str.charAt(j)){
str.delete(i, j+1);
midlen = str.length()/2;
System.out.println("i="+i+",j="+j+ ",len="+ str.length() + ",midlen=" + midlen+ ", after deleted = " + str);
}
}
}
return str.toString();
}

Another possible solution, in case a string is an ASCII string, is to maintain an array of 256 boolean elements to denote ASCII character appearance in a string. If a character appeared for the first time, we keep it and append to the result. Otherwise just skip it.
public String removeDuplicates(String input) {
boolean[] chars = new boolean[256];
StringBuilder resultStringBuilder = new StringBuilder();
for (Character c : input.toCharArray()) {
if (!chars[c]) {
resultStringBuilder.append(c);
chars[c] = true;
}
}
return resultStringBuilder.toString();
}
This approach will also work with Unicode string. You just need to increase chars size.

Solution using JDK7:
public static String removeDuplicateChars(final String str){
if (str == null || str.isEmpty()){
return str;
}
final char[] chArray = str.toCharArray();
final Set<Character> set = new LinkedHashSet<>();
for (char c : chArray) {
set.add(c);
}
final StringBuilder sb = new StringBuilder();
for (Character character : set) {
sb.append(character);
}
return sb.toString();
}

String str = "eamparuthik#gmail.com";
char[] c = str.toCharArray();
String op = "";
for(int i=0; i<=c.length-1; i++){
if(!op.contains(c[i] + ""))
op = op + c[i];
}
System.out.println(op);

public static String removeDuplicateChar(String str){
char charArray[] = str.toCharArray();
StringBuilder stringBuilder= new StringBuilder();
for(int i=0;i<charArray.length;i++){
int index = stringBuilder.toString().indexOf(charArray[i]);
if(index <= -1){
stringBuilder.append(charArray[i]);
}
}
return stringBuilder.toString();
}

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class RemoveDuplicacy
{
public static void main(String args[])throws IOException
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
System.out.print("Enter any word : ");
String s = br.readLine();
int l = s.length();
char ch;
String ans=" ";
for(int i=0; i<l; i++)
{
ch = s.charAt(i);
if(ch!=' ')
ans = ans + ch;
s = s.replace(ch,' '); //Replacing all occurrence of the current character by a space
}
System.out.println("Word after removing duplicate characters : " + ans);
}
}

public static void main(String[] args) {
int i,j;
StringBuffer str=new StringBuffer();
Scanner in = new Scanner(System.in);
System.out.print("Enter string: ");
str.append(in.nextLine());
for (i=0;i<str.length()-1;i++)
{
for (j=1;j<str.length();j++)
{
if (str.charAt(i)==str.charAt(j))
str.deleteCharAt(j);
}
}
System.out.println("Removed String: " + str);
}

This is improvement on solution suggested by #Dave. Here, I am implementing in single loop only.
Let's reuse the return of set.add(T item) method and add it simultaneously in StringBuffer if add is successfull
This is just O(n). No need to make a loop again.
String string = "aabbccdefatafaz";
char[] chars = string.toCharArray();
StringBuilder sb = new StringBuilder();
Set<Character> charSet = new LinkedHashSet<Character>();
for (char c : chars) {
if(charSet.add(c) ){
sb.append(c);
}
}
System.out.println(sb.toString()); // abcdeftz

Simple solution is to iterate through the given string and put each unique character into another string(in this case, a variable result ) if this string doesn't contain that particular character.Finally return result string as output.
Below is working and tested code snippet for removing duplicate characters from the given string which has O(n) time complexity .
private static String removeDuplicate(String s) {
String result="";
for (int i=0 ;i<s.length();i++) {
char ch = s.charAt(i);
if (!result.contains(""+ch)) {
result+=""+ch;
}
}
return result;
}
If the input is madam then output will be mad.
If the input is anagram then output will be angrm
Hope this helps.
Thanks

For the simplicity of the code- I have taken hardcore input, one can take input by using Scanner class also
public class KillDuplicateCharInString {
public static void main(String args[]) {
String str= "aaaabccdde ";
char arr[]= str.toCharArray();
int n = arr.length;
String finalStr="";
for(int i=0;i<n;i++) {
if(i==n-1){
finalStr+=arr[i];
break;
}
if(arr[i]==arr[i+1]) {
continue;
}
else {
finalStr+=arr[i];
}
}
System.out.println(finalStr);
}
}

public static void main (String[] args)
{
Scanner sc = new Scanner(System.in);
String s = sc.next();
String str = "";
char c;
for(int i = 0; i < s.length(); i++)
{
c = s.charAt(i);
str = str + c;
s = s.replace(c, ' ');
if(i == s.length() - 1)
{
System.out.println(str.replaceAll("\\s", ""));
}
}
}

package com.st.removeduplicate;
public class RemoveDuplicate {
public static void main(String[] args) {
String str1="shushil",str2="";
for(int i=0; i<=str1.length()-1;i++) {
int count=0;
for(int j=0;j<=i;j++) {
if(str1.charAt(i)==str1.charAt(j))
count++;
if(count >1)
break;
}
if(count==1)
str2=str2+str1.charAt(i);
}
System.out.println(str2);
}
}