Suppose I enter the following string:
" Hello party"
How could you order only the vowels of that chain, but in a descending way?
to example:
"Halle porty"
In the following way I have tried to solve the problem, but when I run it tells me that I have an ExeptionBounce 5. How could I correct it? I am also told that the responsible line is the following:
if(cad[i].equalsIgnoreCase(vocal[j])
private String[]cad;
private String[]vocal={"a","e","i","o","u"};
private String resp="";
private String aux="":
private String []sort;
private int cont=0;
public String ordenar(String cadena)
{
cad= cadena.split("");
for(int i=0; i<cad.legth; i++)
{
for(int j=i, j<vocal.legth; j++)
{
if(cad[i].equalsIgnoreCase(vocal[j])
{
aux+=cad[i];
sort= aux.Split("");
Arrays.sort(sort);
for(int i=0; i<cad.legth; i++)
{
cad[i] = sort[k];
resp+=cad[i];
}
}
}
return resp;
}
Your code would not even compile - how are you running it ?
To answer your question, here is a piece of code which given a string will return the string with the vowels sorted:
public static boolean isVowel(char c) {
return "AEIOUaeiou".indexOf(c) != -1;
}
public static String sortVowelsOnly(String input) {
if (input == null) {
return input;
}
StringBuilder result = new StringBuilder(input);
//keep track at all positions where we saw a vowel
ArrayList<Integer> vowelPositions = new ArrayList<>();
//keep track of all vowels we saw so far
ArrayList<Character> vowels = new ArrayList<>();
for (int charPosition = 0; charPosition < input.length(); charPosition++) {
char currentChar = input.charAt(charPosition);
if(isVowel(currentChar)) {
vowelPositions.add(charPosition);
vowels.add(currentChar);
}
}
Collections.sort(vowels);
//now just iterate over all positions where we saw a vowel, and populate it with the vowels from the sorted list
for(Integer vowelPosition : vowelPositions) {
char vowel = vowels.remove(0);
result.setCharAt(vowelPosition, vowel);
}
return result.toString();
}
Related
I needed to write a function that takes as input two strings. One is the message I want to write and second are given letters. Letters are ordered randomly.There is no guarantee that each letter occurs a similar number of times .some letters might be missing entirely.
The function should determine if I can write message with the given
letters and it should return true or false accordingly.
I coded it and I think it is very fast, but how can I improve it having in mind the string with letters would be very large while the message would be very short?
Is there a fastest way?
import java.util.HashMap;
import java.util.Map;
import java.util.Random;
public class LetterBowl {
public static void main(String []args){
String message = generateRandomStringUpToThousandChars();
String bowlWithLetters = generateRandomStringUpToThousandChars();
if(canConstructMessage(message, bowlWithLetters)) {
System.out.println("Message '" + message + "' can be constructed with letters from bowl : " + bowlWithLetters);
}
}
public static boolean canConstructMessage(String message, String letters) {
Map<Character,Integer> letterMap = stringToCharacterMap(letters);
char[] messageList = stringToCharacterList(message);
for(char c : messageList) {
if (!containsLetterAndSubtract(c,letterMap))
return false;
}
return true;
}
// checks if map(bowl) contains char andsubtract one char from map(or removes it if it is last one)
public static boolean containsLetterAndSubtract(char c, Map<Character,Integer> letterMap) {
if(letterMap.containsKey(c)) {
if(letterMap.get(c) > 1) {
letterMap.put(c, letterMap.get(c) - 1);
} else {
letterMap.remove(c);
}
return true;
}
return false;
}
public static char[] stringToCharacterList(String message) {
return message.replaceAll(" ", "").toCharArray();
}
public static Map<Character,Integer> stringToCharacterMap(String s) {
Map<Character,Integer> map = new HashMap<Character,Integer>();
for (char c : s.toCharArray()) {
if(map.containsKey(c))
map.put(c, map.get(c) + 1);
else
map.put(c, 1);
}
return map;
}
public static String generateRandomStringUpToThousandChars(){
char[] chars = "abcdefghijklmnopqrstuvwxyz".toCharArray();
StringBuilder sb = new StringBuilder();
Random random = new Random();
for (int i = 0; i < random.nextInt(1000); i++) {
char c = chars[random.nextInt(chars.length)];
sb.append(c);
}
String output = sb.toString();
return output;
};
}
For large bowl size and smaller msg size i found this would be mor efficient :
public static boolean canConstructMessageSorted(String message, String bowlWithLetters) {
int counter = 0;
boolean hasLetter;
//sorting
char[] chars = bowlWithLetters.toCharArray();
Arrays.sort(chars);
String sortedBowl = new String(chars);
//sorting
chars = message.toCharArray();
Arrays.sort(chars);
String sortedMsg = new String(chars);
for (int i = 0; i < sortedMsg.length(); i++) {
hasLetter = false;
for( ; counter < sortedBowl.length() ; counter++) {
if(sortedMsg.charAt(i) == sortedBowl.charAt(counter)) {
hasLetter = true;
break;
}
}
if(!hasLetter) return false;
}
return true;
}
You're operating at O(message.size + letters.size). This is the lowest worst-case time-complexity that I could figure out, on hand. Referring to the fastest way, there's always more you could do. For example, defining the method
public static char[] stringToCharacterList(String message)
and only using it once is technically time-inefficient. You could have simply put that body of code within the canConstructMessage() method, saving another item from being placed on, and taken off of the stack. Although this is such a small fragment of time, when you say fastest, it could be worth talking about.
For every letter in letters, remove 1 copy of it from the message. If the message ends up empty, the answer is "yes":
public static boolean canConstructMessage(String message, String letters) {
for (int i = 0; i < letters.length(); i++)
message = message.replaceFirst("" + letters.charAt(i), "");
return message.isEmpty();
}
If reusing letters is allowed, you can do it in 1 line:
public static boolean canConstructMessage(String message, String letters) {
return letters.chars().boxed().collect(Collectors.toSet())
.containsAll(message.chars().boxed().collect(Collectors.toSet());
}
I found this would be more efficient for large bowl size and small msg size :
public static boolean canConstructMessageSorted(String message, String bowlWithLetters) {
int counter = 0;
boolean hasLetter;
//sorting
char[] chars = bowlWithLetters.toCharArray();
Arrays.sort(chars);
String sortedBowl = new String(chars);
//sorting
chars = message.toCharArray();
Arrays.sort(chars);
String sortedMsg = new String(chars);
for (int i = 0; i < sortedMsg.length(); i++) {
hasLetter = false;
for( ; counter < sortedBowl.length() ; counter++) {
if(sortedMsg.charAt(i) == sortedBowl.charAt(counter)) {
hasLetter = true;
break;
}
}
if(!hasLetter) return false;
}
return true;
}
I am having difficulties with my method returning true. It is a boolean method that takes two words and tries to see if one can be turned into the other by transposing two neighboring letters. I have had no troubles getting the false boolean. When the code gets to the for loop with an if statement in it it runs fine but does not return true when the if statement is satisfied. For some reason it continues through the for loop. For example, when comparing "teh" and "the" when the loop hits 1 the if statement is satisfied but does not return true, the for lo
public static boolean transposable(String word1, String word2)
{
ArrayList<Character> word1char = new ArrayList<Character>();
ArrayList<Character> word2char = new ArrayList<Character>();
int word1length = word1.length();
int word2length = word2.length();
int count = 0;
String w1 = word1.toUpperCase();
String w2 = word2.toUpperCase();
if(word1length != word2length)
{
return false;
}
for(int i = 0; i < word1length; i++)
{
char letter1 = w1.charAt(i);
word1char.add(letter1);
char letter2 = w2.charAt(i);
word2char.add(letter2);
}
for(int i = 0; i < word1length; i++)
{
char w1c = word1char.get(i);
char w2c = word2char.get(i);
if(w1c == w2c)
{
count++;
}
}
if(count < word1length - 2)
{
return false;
}
for(int i = 0; i < word1length; i++)
{
char w1c = word1char.get(i);
char w2c = word2char.get(i+1);
if(w1c == w2c)
{
return true;
}
}
return false;
}
op just keeps running. What am I doing wrong?
As pointed out in the comments this doesn't seem to be the easiest way around this problem. Here is a solution which tries to follow your logic and includes the use of toUpperCase() and ArrayLists.
Going over your code it looks like you were getting a bit lost in your logic. This is because you had one method trying to do everything. Break things down into smaller methods and you also will benefit by not having to repeat code and it keeps things much cleaner. The code below is tested with Java8 (although there is no reason why this should not work with Java 7).
public static void main(String args[]) {
String word1 = "Hello";
String word2 = "Hlelo";
transposable(word1, word2);
}
private static boolean transposable(String word1, String word2) {
// Get an ArrayList of characters for both words.
ArrayList<Character> word1CharacterList = listOfCharacters(word1);
ArrayList<Character> word2CharacterList = listOfCharacters(word2);
boolean areWordsEqual;
// Check that the size of the CharacterLists is the same
if (word1CharacterList.size() != word2CharacterList.size()) {
return false;
}
// check to see if words are equal to start with
areWordsEqual = checkIfTwoWordsAreTheSame(word1CharacterList, word2CharacterList);
System.out.print("\n" + "Words are equal to be begin with = " + areWordsEqual);
if (!areWordsEqual) {
/*
This loop i must start at 1 because you can't shift an ArrayList index of 0 to the left!
Loops through all the possible combinations and checks if there is a match.
*/
for (int i = 1; i < word1CharacterList.size(); i++) {
ArrayList<Character> adjustedArrayList = shiftNeighbouringCharacter(word2CharacterList, i);
areWordsEqual = checkIfTwoWordsAreTheSame(word1CharacterList, adjustedArrayList);
System.out.print("\n" + "Loop count " + i + " words are equal " + areWordsEqual + word1CharacterList + adjustedArrayList.toString());
if (areWordsEqual) {
break;
}
}
}
return areWordsEqual;
}
// takes in a String as a parameter and returns an ArrayList of Characters in the order of the String parameter.
private static ArrayList<Character> listOfCharacters(String word) {
ArrayList<Character> wordCharacters = new ArrayList<Character>();
String tempWord = word.toUpperCase();
for (int wordLength = 0; wordLength < tempWord.length(); wordLength++) {
Character currentCharacter = tempWord.charAt(wordLength);
wordCharacters.add(currentCharacter);
}
return wordCharacters;
}
// takes in two character arrayLists, and compares each index character.
private static boolean checkIfTwoWordsAreTheSame(ArrayList<Character> characterList1, ArrayList<Character> characterList2) {
// compare list1 against list two
for (int i = 0; i < characterList1.size(); i++) {
Character currentCharacterList1 = characterList1.get(i);
Character currentCharacterList2 = characterList2.get(i);
if (!currentCharacterList1.equals(currentCharacterList2)) {
return false;
}
}
return true;
}
// this method takes in an ArrayList of characters and the initial index that we want to shift one place to the left.
private static ArrayList<Character> shiftNeighbouringCharacter(ArrayList<Character> characterListToShift, int indexToShiftLeft) {
ArrayList<Character> tempCharacterList = new ArrayList<Character>();
int indexAtLeft = indexToShiftLeft - 1;
// fill the new arrayList full of nulls. We will have to remove these nulls later before we can add proper values in their place.
for (int i = 0; i < characterListToShift.size(); i++) {
tempCharacterList.add(null);
}
//get the current index of indexToShift
Character characterOfIndexToShift = characterListToShift.get(indexToShiftLeft);
Character currentCharacterInThePositionToShiftTo = characterListToShift.get(indexAtLeft);
tempCharacterList.remove(indexAtLeft);
tempCharacterList.add(indexAtLeft, characterOfIndexToShift);
tempCharacterList.remove(indexToShiftLeft);
tempCharacterList.add(indexToShiftLeft, currentCharacterInThePositionToShiftTo);
for (int i = 0; i < characterListToShift.size(); i++) {
if (tempCharacterList.get(i) == null) {
Character character = characterListToShift.get(i);
tempCharacterList.remove(i);
tempCharacterList.add(i, character);
}
}
return tempCharacterList;
}
Hope this helps. If you are still struggling then follow along in your debugger. :)
So my task is to write a program that counts the number of words and unique words in a given string that we get from the user without using arrays.
I can do the first task and was wondering how I could go about doing the second part.
For counting the number of words in the string I have
boolean increment = false;
for (int i = 0; i < inputPhrase.length(); i++){
if(validChar(inputPhrase.charAt(i))) //validChar(char c) is a simple method that returns a valid character{
increment = true;
}
else if(increment){
phraseWordCount ++;
increment = false;
}
}
if(increment) phraseWordCount++; //in the case the last word is a valid character
(originally i left this out and was off by one word)
to count unique words can I somehow modify this?
Here a suggestion how to do it without arrays:
1) Read every char until a blank is found and add this char to a second String.
2) If a blank is found, add it (or another token to seperate words) to the second String.
2a) Read every word from second String comparing it to the current word from he input String
public static void main(String[] args) {
final String input = "This is a sentence that is containing three times the word is";
final char token = '#';
String processedInput = "";
String currentWord = "";
int wordCount = 0;
int uniqueWordCount = 0;
for (char c : input.toCharArray()) {
if (c != ' ') {
processedInput += c;
currentWord += c;
} else {
processedInput += token;
wordCount++;
String existingWord = "";
int occurences = 0;
for (char c1 : processedInput.toCharArray()) {
if (c1 != token) {
existingWord += c1;
} else {
if (existingWord.equals(currentWord)) {
occurences++;
}
existingWord = "";
}
}
if (occurences <= 1) {
System.out.printf("New word: %s\n", currentWord);
uniqueWordCount++;
}
currentWord = "";
}
}
wordCount++;
System.out.printf("%d words total, %d unique\n", wordCount, uniqueWordCount);
}
Output
New word: This
New word: is
New word: a
New word: sentence
New word: that
New word: containing
New word: three
New word: times
New word: the
New word: word
12 words total, 10 unique
Using the Collections API you can count words with the following method:
private int countWords(final String text) {
Scanner scanner = new Scanner(text);
Set<String> uniqueWords = new HashSet<String>();
while (scanner.hasNext()) {
uniqueWords.add(scanner.next());
}
scanner.close();
return uniqueWords.size();
}
If it is possible that you get normal sentences with punctuation marks you can change the second line to:
Scanner scanner = new Scanner(text.replaceAll("[^0-9a-zA-Z\\s]", "").toLowerCase());
Every time a word ends findUpTo checks if the word is contained in the input before the start of that word. So "if if if" would count as one unique and three total words.
/**
* Created for http://stackoverflow.com/q/22981210/1266906
*/
public class UniqueWords {
public static void main(String[] args) {
String inputPhrase = "one two ones two three one";
countWords(inputPhrase);
}
private static void countWords(String inputPhrase) {
boolean increment = false;
int wordStart = -1;
int phraseWordCount = 0;
int uniqueWordCount = 0;
for (int i = 0; i < inputPhrase.length(); i++){
if(validChar(inputPhrase.charAt(i))) { //validChar(char c) is a simple method that returns a valid character{
increment = true;
if(wordStart == -1) {
wordStart = i;
}
} else if(increment) {
phraseWordCount++;
final String lastWord = inputPhrase.substring(wordStart, i);
boolean unique = findUpTo(lastWord, inputPhrase, wordStart);
if(unique) {
uniqueWordCount++;
}
increment = false;
wordStart = -1;
}
}
if(increment) {
phraseWordCount++; //in the case the last word is a valid character
final String lastWord = inputPhrase.substring(wordStart, inputPhrase.length());
boolean unique = findUpTo(lastWord, inputPhrase, wordStart);
if(unique) {
uniqueWordCount++;
}
}
System.out.println("Words: "+phraseWordCount);
System.out.println("Unique: "+uniqueWordCount);
}
private static boolean findUpTo(String needle, String haystack, int lastPos) {
boolean previousValid = false;
boolean unique = true;
for(int j = 0; unique && j < lastPos - needle.length(); j++) {
final boolean nextValid = validChar(haystack.charAt(j));
if(!previousValid && nextValid) {
// Word start
previousValid = true;
for (int k = 0; k < lastPos - j; k++) {
if(k == needle.length()) {
// We matched all characters. Only if the word isn't finished it is unique
unique = validChar(haystack.charAt(j+k));
break;
}
if (needle.charAt(k) != haystack.charAt(j+k)) {
break;
}
}
} else {
previousValid = nextValid;
}
}
return unique;
}
private static boolean validChar(char c) {
return Character.isAlphabetic(c);
}
}
My code is throwing an error that I've never seen before. So hey! I guess I'm learning ;) Anyway, I did some reading and generally this error is thrown when a list that is being iterated over is modified mid-iteration. However, I'm pretty sure I'm not modifying it. While the error is being thrown on partition(), if I don't assign a new value for currentList in updateCurrentList() (by commenting out the code), the program no longer throws the error. These two functions are called one after the other in my play() method, however the list iteration should be complete by the time the change is made. What am I missing? Do I have to close down the iterator somehow?
package hangman;
import java.io.*;
import java.util.*;
import javax.swing.JOptionPane;
public class Hangman {
private Map<String, List<String>> wordPartitions; // groups words according to positions of guessed letter
private List<String> currentList; // remaining possible words that fit the information given so far
Set<Character> wrongGuesses; // holds all the "wrong" guesses so far
StringBuilder guessString; // current state of the word being guessed
String justHyphens; // for checking whether a guess was "wrong"
// initialize fields
// currentList should contain all (and only) words of length wordLength
// justHyphens and guessString should consist of wordLength hyphens
public Hangman(int wordLength) throws FileNotFoundException {
this.currentList = new ArrayList<String>();
addWords(wordLength);
wrongGuesses = new HashSet();
for(int i = 0; i < wordLength; i++) {
justHyphens += "-";
}
guessString = new StringBuilder();
wordPartitions = new HashMap();
}
private void addWords(int wordLength) throws FileNotFoundException {
Scanner words = new Scanner(new File("lexicon.txt"));
String word = "";
while(words.hasNext()) {
word = words.next();
if (word.length() == wordLength) {
currentList.add(word);
}
}
}
// main loop
public void play() {
char choice;
do {
choice = getUserChoice();
partition(choice);
updateCurrentList(choice);
} while (!gameOver());
endMessage();
}
// display the guessString and the missed guesses
// and get the next guess
private char getUserChoice() {
//generate a string from the incorrect choices char list
String wrong = "";
char letter;
if(!wrongGuesses.isEmpty()) {
Iterator<Character> letters = wrongGuesses.iterator();
letter = letters.next();
while(letters.hasNext()) {
letter = letters.next();
wrong += ", " + letter;
}
}
String letterStr = JOptionPane.showInputDialog("Incorrect choices: "+ wrong +"\n Tested letters: "+ guessString.toString() +"\nplease input a letter.");
return letterStr.charAt(0);
}
// use wordPartitions to partition currentList using
// keys returned by getPartitionKey()
private void partition(char choice) {
String word = "";
String key = "";
List<String> tempList = new ArrayList<String>();
Iterator<String> words = currentList.iterator();
//Generate a key for each word and add to appropriate arraylist within map.
while(words.hasNext()) {
word = words.next();
key = getPartitionKey(word, choice);
if(wordPartitions.containsKey(key)) {
tempList = wordPartitions.get(key);
tempList.add(word);
wordPartitions.put(key, tempList);
} else {
tempList.clear();
tempList.add(word);
wordPartitions.put(key, new ArrayList<String>());
}
}
}
// update currentList to be a copy of the longest partition
// if choice was "wrong", add choice to wrongGuesses
// if choice was "right", update guessString
private void updateCurrentList(char choice) {
String key = findLongestList();
currentList = wordPartitions.get(key);
if(key.equals(justHyphens)) {
wrongGuesses.add(choice);
} else {
addLetterToGuessString(guessString, choice, key);
}
}
private String findLongestList() {
Set<String> keySet = wordPartitions.keySet();
Iterator<String> keys = keySet.iterator();
String maxKey = "";
int maxKeyLength = 0;
List<String> tempList;
String tempKey = "";
while(keys.hasNext()) {
tempKey = keys.next();
tempList = wordPartitions.get(tempKey);
if(tempList.size() > maxKeyLength) {
maxKeyLength = tempList.size();
maxKey = tempKey;
}
}
return maxKey;
}
// checks for end of game
private boolean gameOver() {
return false;
}
// display the guessString and the missed guesses
// and print "Congratulations!"
private void endMessage() {
JOptionPane.showMessageDialog(null, "Congrats, yo!");
}
// returns string with '-' in place of each
// letter that is NOT the guessed letter
private String getPartitionKey(String s, char c) {
String word = "";
String letter = Character.toString(c);
for(int i = 0; i < s.length(); i++) {
if(s.charAt(i) == c) {
word += letter;
} else {
word += "-";
}
}
return word;
}
// update guessString with the guessed letter
private void addLetterToGuessString(StringBuilder guessString, char letter, String key) {
for(int i = 0; i < key.length(); i++) {
if(key.charAt(i) != '-') {
guessString.setCharAt(i, key.charAt(i));
}
}
}
}
The problem is that you are modifying a collection while you are iterating over it.
The collection is currentList, you are iterating over it in partition(). You modify it when you add a word to tempList here:
key = getPartitionKey(word, choice);
if(wordPartitions.containsKey(key)) {
tempList = wordPartitions.get(key);
tempList.add(word);
wordPartitions.put(key, tempList);
} else {
Why ? Because previously you called updateCurrentList() from play():
do {
choice = getUserChoice();
partition(choice);
updateCurrentList(choice);
} while (!gameOver());
And you updated currentList:
String key = findLongestList();
currentList = wordPartitions.get(key);
So, if the key returned by getPartitionKey(word, choice) is the same as the key previously returned by findLongestList(), currentListwill be the same as tempList, and so you will be modifying the collection you are iterating over.
The solution ? If tempList is the same as currentList, don't add the word to it (it already have the word, by definition). So, you can rewrite your if-else like that (I removed some useless code):
if(wordPartitions.containsKey(key)) {
tempList = wordPartitions.get(key);
} else {
wordPartitions.put(key, new ArrayList<String>());
}
if (tempList!=currentList) {
tempList.add(word);
}
In my program, the user enters a string, and it first finds the largest mode of characters in the string. Next, my program is supposed to remove all duplicates of a character in a string, (user input: aabc, program prints: abc) which I'm not entirely certain on how to do. I can get it to remove duplicates from some strings, but not all. For example, when the user puts "aabc" it will print "abc", but if the user puts "aabbhh", it will print "abbhh." Also, before I added the removeDup method to my program, it would only print the maxMode once, but after I added the removeDup method, it began to print the maxMode twice. How do I keep it from printing it twice?
Note: I cannot convert the strings to an array.
import java.util.Scanner;
public class JavaApplication3 {
static class MyStrings {
String s;
void setMyStrings(String str) {
s = str;
}
int getMode() {
int i;
int j;
int count = 0;
int maxMode = 0, maxCount = 1;
for (i = 0; i< s.length(); i++) {
maxCount = count;
count = 0;
for (j = s.length()-1; j >= 0; j--) {
if (s.charAt(j) == s.charAt(i))
count++;
if (count > maxCount){
maxCount = count;
maxMode = i;
}
}
}
System.out.println(s.charAt(maxMode)+" = largest mode");
return maxMode;
}
String removeDup() {
getMode();
int i;
int j;
String rdup = "";
for (i = 0; i< s.length(); i++) {
int count = 1;
for (j = 0; j < rdup.length(); j++) {
if (s.charAt(i) == s.charAt(j)){
count++;
}
}
if (count == 1){
rdup += s.charAt(i);
}
}
System.out.print(rdup);
System.out.println();
return rdup;
}
}
public static void main (String[] args) {
Scanner in = new Scanner(System.in);
MyStrings setS = new MyStrings();
String s;
System.out.print("Enter string:");
s = in.nextLine();
setS.setMyStrings(s);
setS.getMode();
setS.removeDup();
}
}
Try this method...should work fine!
String removeDup()
{
getMode();
int i;
int j;
String rdup = "";
for (i = 0; i< s.length(); i++) {
int count = 1;
for (j = i+1; j < s.length(); j++) {
if (s.charAt(i) == s.charAt(j)) {
count++;
}
}
if (count == 1){
rdup += s.charAt(i);
}
}
// System.out.print(rdup);
System.out.println();
return rdup;
}
Welcome to StackOverflow!
You're calling getMode() both outside and inside of removeDup(), which is why it's printing it twice.
In order to remove all duplicates, you'll have to call removeDup() over and over until all the duplicates are gone from your string. Right now you're only calling it once.
How might you do that? Think about how you're detecting duplicates, and use that as the end condition for a while loop or similar.
Happy coding!
Shouldn't this be an easier way? Also, i'm still learning.
import java.util.*;
public class First {
public static void main(String arg[])
{
Scanner sc= new Scanner(System.in);
StringBuilder s=new StringBuilder(sc.nextLine());
//String s=new String();
for(int i=0;i<s.length();i++){
String a=s.substring(i, i+1);
while(s.indexOf(a)!=s.lastIndexOf(a)){s.deleteCharAt(s.lastIndexOf(a));}
}
System.out.println(s.toString());
}
}
You can do this:
public static void main(String[] args) {
String str = new String("PINEAPPLE");
Set <Character> letters = new <Character>HashSet();
for (int i = 0; i < str.length(); i++) {
letters.add(str.charAt(i));
}
System.out.println(letters);
}
I think an optimized version which supports ASCII codes can be like this:
public static void main(String[] args) {
System.out.println(removeDups("*PqQpa abbBBaaAAzzK zUyz112235KKIIppP!!QpP^^*Www5W38".toCharArray()));
}
public static String removeDups(char []input){
long ocr1=0l,ocr2=0l,ocr3=0;
int index=0;
for(int i=0;i<input.length;i++){
int val=input[i]-(char)0;
long ocr=val<126?val<63?ocr1:ocr2:ocr3;
if((ocr& (1l<<val))==0){//not duplicate
input[index]=input[i];
index++;
}
if(val<63)
ocr1|=(1l<<val);
else if(val<126)
ocr2|=(1l<<val);
else
ocr3|=(1l<<val);
}
return new String(input,0,index);
}
please keep in mind that each of orc(s) represent a mapping of a range of ASCII characters and each java long variable can grow as big as (2^63) and since we have 128 characters in ASCII so we need three ocr(s) which basically maps the occurrences of the character to a long number.
ocr1: (char)0 to (char)62
ocr2: (char)63 to (char)125
ocr3: (char)126 to (char)128
Now if a duplicate was found the
(ocr& (1l<<val))
will be greater than zero and we skip that char and finally we can create a new string with the size of index which shows last non duplicate items index.
You can define more orc(s) and support other character-sets if you want.
Can use HashSet as well as normal for loops:
public class RemoveDupliBuffer
{
public static String checkDuplicateNoHash(String myStr)
{
if(myStr == null)
return null;
if(myStr.length() <= 1)
return myStr;
char[] myStrChar = myStr.toCharArray();
HashSet myHash = new HashSet(myStrChar.length);
myStr = "";
for(int i=0; i < myStrChar.length ; i++)
{
if(! myHash.add(myStrChar[i]))
{
}else{
myStr += myStrChar[i];
}
}
return myStr;
}
public static String checkDuplicateNo(String myStr)
{
// null check
if (myStr == null)
return null;
if (myStr.length() <= 1)
return myStr;
char[] myChar = myStr.toCharArray();
myStr = "";
int tail = 0;
int j = 0;
for (int i = 0; i < myChar.length; i++)
{
for (j = 0; j < tail; j++)
{
if (myChar[i] == myChar[j])
{
break;
}
}
if (j == tail)
{
myStr += myChar[i];
tail++;
}
}
return myStr;
}
public static void main(String[] args) {
String myStr = "This is your String";
myStr = checkDuplicateNo(myStr);
System.out.println(myStr);
}
Try this simple answer- works well for simple character string accepted as user input:
import java.util.Scanner;
public class string_duplicate_char {
String final_string = "";
public void inputString() {
//accept string input from user
Scanner user_input = new Scanner(System.in);
System.out.println("Enter a String to remove duplicate Characters : \t");
String input = user_input.next();
user_input.close();
//convert string to char array
char[] StringArray = input.toCharArray();
int StringArray_length = StringArray.length;
if (StringArray_length < 2) {
System.out.println("\nThe string with no duplicates is: "
+ StringArray[1] + "\n");
} else {
//iterate over all elements in the array
for (int i = 0; i < StringArray_length; i++) {
for (int j = i + 1; j < StringArray_length; j++) {
if (StringArray[i] == StringArray[j]) {
int temp = j;//set duplicate element index
//delete the duplicate element by copying the adjacent elements by one place
for (int k = temp; k < StringArray_length - 1; k++) {
StringArray[k] = StringArray[k + 1];
}
j++;
StringArray_length--;//reduce char array length
}
}
}
}
System.out.println("\nThe string with no duplicates is: \t");
//print the resultant string with no duplicates
for (int x = 0; x < StringArray_length; x++) {
String temp= new StringBuilder().append(StringArray[x]).toString();
final_string=final_string+temp;
}
System.out.println(final_string);
}
public static void main(String args[]) {
string_duplicate_char object = new string_duplicate_char();
object.inputString();
}
}
Another easy solution to clip the duplicate elements in a string using HashSet and ArrayList :
import java.util.ArrayList;
import java.util.HashSet;
import java.util.Iterator;
import java.util.Scanner;
public class sample_work {
public static void main(String args[]) {
String input = "";
System.out.println("Enter string to remove duplicates: \t");
Scanner in = new Scanner(System.in);
input = in.next();
in.close();
ArrayList<Character> String_array = new ArrayList<Character>();
for (char element : input.toCharArray()) {
String_array.add(element);
}
HashSet<Character> charset = new HashSet<Character>();
int array_len = String_array.size();
System.out.println("\nLength of array = " + array_len);
if (String_array != null && array_len > 0) {
Iterator<Character> itr = String_array.iterator();
while (itr.hasNext()) {
Character c = (Character) itr.next();
if (charset.add(c)) {
} else {
itr.remove();
array_len--;
}
}
}
System.out.println("\nThe new string with no duplicates: \t");
for (int i = 0; i < array_len; i++) {
System.out.println(String_array.get(i).toString());
}
}
}
your can use this simple code and understand how to remove duplicates values from string.I think this is the simplest way to understand this problem.
class RemoveDup
{
static int l;
public String dup(String str)
{
l=str.length();
System.out.println("length"+l);
char[] c=str.toCharArray();
for(int i=0;i<l;i++)
{
for(int j=0;j<l;j++)
{
if(i!=j)
{
if(c[i]==c[j])
{
l--;
for(int k=j;k<l;k++)
{
c[k]=c[k+1];
}
j--;
}
}
}
}
System.out.println("after concatination lenght:"+l);
StringBuilder sd=new StringBuilder();
for(int i=0;i<l;i++)
{
sd.append(c[i]);
}
str=sd.toString();
return str;
}
public static void main(String[] ar)
{
RemoveDup obj=new RemoveDup();
Scanner sc=new Scanner(System.in);
String st,t;
System.out.println("enter name:");
st=sc.nextLine();
sc.close();
t=obj.dup(st);
System.out.println(t);
}
}
/*
* To change this license header, choose License Headers in Project Properties.
* To change this template file, choose Tools | Templates
* and open the template in the editor.
*/
package javaapplication26;
import java.util.*;
/**
*
* #author THENNARASU
*/
public class JavaApplication26 {
public static void main(String[] args) {
int i,j,k=0,count=0,m;
char a[]=new char[10];
char b[]=new char[10];
Scanner ob=new Scanner(System.in);
String str;
str=ob.next();
a=str.toCharArray();
int c=str.length();
for(j=0;j<c;j++)
{
for(i=0;i<j;i++)
{
if(a[i]==a[j])
{
count=1;
}
}
if(count==0)
{
b[k++]=a[i];
}
count=0;
}
for(m=0;b[m]!='\0';m++)
{
System.out.println(b[m]);
}
}
}
i wrote this program. Am using 2 char arrays instead. You can define the number of duplicate chars you want to eliminate from the original string and also shows the number of occurances of each character in the string.
public String removeMultipleOcuranceOfChar(String string, int numberOfChars){
char[] word1 = string.toCharArray();
char[] word2 = string.toCharArray();
int count=0;
StringBuilder builderNoDups = new StringBuilder();
StringBuilder builderDups = new StringBuilder();
for(char x: word1){
for(char y : word2){
if (x==y){
count++;
}//end if
}//end inner loop
System.out.println(x + " occurance: " + count );
if (count ==numberOfChars){
builderNoDups.append(x);
}else{
builderDups.append(x);
}//end if else
count = 0;
}//end outer loop
return String.format("Number of identical chars to be in or out of input string: "
+ "%d\nOriginal word: %s\nWith only %d identical chars: %s\n"
+ "without %d identical chars: %s",
numberOfChars,string,numberOfChars, builderNoDups.toString(),numberOfChars,builderDups.toString());
}
Try this simple solution for REMOVING DUPLICATE CHARACTERS/LETTERS FROM GIVEN STRING
import java.util.Scanner;
public class RemoveDuplicateLetters {
public static void main(String[] args) {
Scanner scn=new Scanner(System.in);
System.out.println("enter a String:");
String s=scn.nextLine();
String ans="";
while(s.length()>0)
{
char ch = s.charAt(0);
ans+= ch;
s = s.replace(ch+"",""); //Replacing all occurrence of the current character by a spaces
}
System.out.println("after removing all duplicate letters:"+ans);
}
}
In Java 8 we can do that using
private void removeduplicatecharactersfromstring() {
String myString = "aabcd eeffff ghjkjkl";
StringBuilder builder = new StringBuilder();
Arrays.asList(myString.split(" "))
.forEach(s -> {
builder.append(Stream.of(s.split(""))
.distinct().collect(Collectors.joining()).concat(" "));
});
System.out.println(builder); // abcd ef ghjkl
}