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Converting array list to a string array

I've been trying to convert my string array list to a string array so I can print it but have been unable to do so.
This is the class I have, randomQuestion which takes in an array list from the gameQuestions method in the same class.
I have never tried to convert an array list using a loop before hence the difficulty, I was able to convert it fine with the code
String[] questions = data1.toArray(new String[]{});
But I need it to loop through using a for loop to store it in an array which I can then print one at a time once a question is answered successfully.
The error I'm receiving from netbeans is cannot find symbol
Symbol:methodtoArray(String[]) for the .toArray portion below.
public String[] randomQuestion(ArrayList data1) {
Collections.shuffle(data1);
for (int question = 0; question < 10; question++) {
ranquestions = data1.get(question).toArray(new String[10]);
}
return ranquestions;
}
Any help would be greatly appreciated.

You can use List.toArray(). Class List has a method:
<T> T[] toArray(T[] a);

Assuming you have an ArrayList<String>, you can use String.join(delimiter, wordList) in order to concatenate all the elements to a single String:
public static void main(String[] args) {
// example list
List<String> words = new ArrayList<String>();
words.add("You");
words.add("can");
words.add("concatenate");
words.add("these");
words.add("Strings");
words.add("in");
words.add("one");
words.add("line");
// concatenate the elements delimited by a whitespace
String sentence = String.join(" ", words);
// print the result
System.out.println(sentence);
}
The result of this example is
You can concatenate these Strings in one line
So using your list, String.join(" ", data1) would create a String with the elements of data1 delimited by a whitespace.

The question is how to create an array with only 10 elements of the list, if I understood correctly.
Streams (Java 8):
String[] ranquestions = data1.stream()
.limit(10)
.toArray(String[]::new);
Loop (based on question, avoiding unnecessary changes):
String[] ranquestions = new String[10];
for(int question = 0; question < 10; question++) {
ranquestions[question] = data1.get(question);
}
always assuming List<String> data1, if not some conversion is needed.
Example:
String[] ranquestions = data1.stream()
.limit(10)
.map(String::valueOf)
.toArray(String[]::new);
or, loop case:
ranquestions[question] = String.valueOf(data1.get(question));

You can do:
private String[] randomQuestions(ArrayList data){
Collections.shuffle(data);
return (String[]) data.toArray();
}
If you are sure you are getting a list of string (question) you can instead
private String[] randomQuestions(List<String> data){
Collections.shuffle(data);
return (String[]) data.toArray();
}
Edit 1
private static String[] randomQuestions(ArrayList data){
Collections.shuffle(data);
String[] randomQuestions = new String[data.size()];
for(int i=0; i<data.size(); i++){
randomQuestions[i] = String.valueOf(data.get(i));
}
return randomQuestions;
}

Create a Set of lists from two lists through recursion

I've searched through many questions on this site with somewhat similar underlying concepts, however after many hours of attempting to solve this problem myself and reviewing I am still lost. If there is another question that answers this I will be more than happy to give it a look over.
Ultimately I want to create a recursive method such that it takes two lists and returns a Set of String lists:
//Example of such a function definition
private static Set<List<String>> myRecursiveMethod(List<String> listOne,
List<String> listTwo) {
}
When I say "Set of String lists" I mean specifically the following:
(Note:"AD" == "DA")
// if the two following lists are INPUTTED into myRecursiveMethod();
// listOne = ["A","B"]
// listTwo = ["C","D"]
// the following set is OUTPUTTED: [["AC","BD"],["AD","BC"]]
Such that if there were three elements in both listOne and listTwo, there would be SIX elements in the set. i.e:
// listOne = ["A","B","C"]
// listTwo = ["D","E","F"]
// OUTPUTTED: [["AD","BE","CF"],["AD","BF","CE"],["BD","AE","CF"],
// ["BD","AF","CE"],["CD","AE","BF"],["CD","AF","BE"]]
I tried writing this using a double enhanced FOR loop so I could understand the logic. My FOR loop approach is terrible and only works for the HARD-CODED limit of list.size() == 2.
// Create Lists and append elements
List<String> listOne = new ArrayList<String>();
listOne.add("A");
listOne.add("B");
List<String> listTwo = new ArrayList<String>();
listTwo.add("C");
listTwo.add("D");
// List One = ["A","B"]
// List Two = ["C","D"]
// Create new List
List<List<String>> newList = new ArrayList<List<String>>();
Integer counter = 0;
for (String s : listOne) {
counter++;
for (String p : listTwo) {
// A HARD-CODED bad implementation of this method
if (counter < 3) {
List<String> newListTwo = new ArrayList<String>();
newListTwo.add(s.concat(p));
newList.add(newListTwo);
} else if (!(counter % 2 == 0)) {
newList.get(1).add(s.concat(p));
} else {
newList.get(0).add(s.concat(p));
}
}
}
System.out.println(newList); // = [["AC","BD"],["AD","BC"]]
Also you can note that I defined List<List<String>> Rather than Set<List<String>>. This was due to my badly coded attempted which relies on the list.get() method.
So my current recursive method is as follows:
private static Set<List<String>> myRecursiveMethod(List<String> listOne,
List<String> listTwo)
{
//Base Case:
if (listOne.isEmpty){
return new HashSet<List<String>>;
}
//Recursive Case:
else {
String listOneFirst = listOne.get(0);
String listTwoFirst = listTwo.get(0);
List<String> sampleList = new ArrayList<String>();
sampleList.add(listOneFirst+listTwoFirst);
Set<List<String>> newSet = new HashSet<List<String>>(myRecursiveMethod())
newSet.add(sampleList);
return newSet;
}
}
This method only acts like this currently:
INPUT:
List One = ["A","B"]
List Two = ["C","D"]
OUTPUT:
[["AC"]["BD"]]
DESIRED OUTPUT:
[["AC","BD"],["AD","BC"]]
EDIT:
After reviewing responses my W.I.P code for the class:
private static Set<List<String>> myRecursiveMethod(List<String> listOne,
List<String> listTwo) {
//Backup Case (user enters an empty list)
if (listOne.isEmpty()){
return new HashSet<List<String>>();
}
// Base Case:
if (listOne.size() == 1) {
List<String> mergedStrings = new ArrayList<>();
for (String s : listTwo) {
mergedStrings.add(listOne.get(0).concat(s));
}
Set<List<String>> builtHashSet = new HashSet<List<String>();
builtHashSet.add(mergedStrings);
return builtHashSet;
}
// Recursive Case:
else {
// Ensure original list values arn't changed.
List<String> newListOne = new ArrayList<String>(listOne);
List<String> newListTwo = new ArrayList<String>(listTwo);
//first two elements...I don't think this is correct
String listOneFirst = newListOne.get(0);
String listTwoFirst = newListTwo.get(0);
List<String> sampleList = new ArrayList<String>();
sampleList.add(listOneFirst + listTwoFirst);
//used for making recursive case smaller
newListOne.remove(0);
// Calls recursion
Set<List<String>> newSet = new HashSet<List<String>>(
myRecursiveMethod(newListOne, newListTwo));
newSet.add(sampleList);
return newSet;
}
}

I think the problem is here:
if (listOne.isEmpty){
return new HashSet<List<String>>;
}
You are correct, at some point your recursion has to end, and you have to start building the desired output. But the desired output is not a Set with an empty list. It is a Set containing some lists with some content. Thus: don't wait until listOne is empty. Instead:
if (listOne.size() == 1) {
List<String> mergedStrings = new ArrayList<>();
mergedStrings = ... merge the ONE listOne entry with all listTwo entries
Set<List<String>> rv = new HashSet<>();
rv.add(mergedStrings);
return rv;
}
In other words: you use recursion to reduce the length of the first list by one. And when only one element is left in that list, it is time to merge in the second list.
Now lets look into how to "use" that (calling the method rec for brevity); putting down some pseudo code to show the steps we need:
rec([a, b], [c,d]) -->
rec([a], [c,d]) X rec([b], [c, d]) -->
<[ac, ad]> X <[bc, bd]> -->
<[ac, ad], [bc, bd]>
"X" meaning "joining" two results from recursive calls; should be as easy as:
Set<List<String>> rec1 = rec(...);
return rec1.addAll(rec2 ...

How would I get all the words from a list that begins with the specified letter

I am trying to show the list of words which start with the letter specified by the user input.
So for example if I add three words to my list, cat, corn and dog, and the user inputs the letter c, the output on the Java applet should be cat, corn.
However, I have no idea on how to go about this.
public void actionPerformed(ActionEvent e){
if (e.getSource() == b1 ){
x = textf.getText();
wordList.add(x);
textf.setText(null);
}
if (e.getSource() == b2 ){
}
}
b1 is adding all the user input into a secretly stored list, and I now want to make another button when pressed to show the words that start with the specified letter by the user.
textf = my text field
wordList = my list I created
x = string I previously defined

You could loop through all the possible indices, check if the element at that index starts with the letter, and print it if it does.
ALTERNATIVE (and probably better) code (I was going to put this after, but since its better it deserves to be first. Taken form #larsmans's answer here.
//given wordList as the word list
//given startChar as the character to search for in the form of a *String* not char
for (String element : wordList){
if (element.startsWith(startChar)){
System.out.println(element);
}
}
DISCLAIMER: This code is untested, I don't have much experience with ArrayList, and Java is more of a quaternary programming language for me. Hope it works :)
//given same variables as before
for (int i = 0; i < wordList.size(); i++){
String element = wordList.get(i);
//you could remove the temporary variable and replace element with
// wordList.get(i)
if (element.startsWith(startChar){
System.out.println(element);
}
}

You can try something like this -
public static void main(String[] args) {
String prefix = "a";
List<String> l = new ArrayList<String>();
List<String> result = new ArrayList<String>();
l.add("aah");
l.add("abh");
l.add("bah");
for(String s: l) {
if(s.startsWith(prefix)) {
result.add(s);
}
}
System.out.println(result);
}
Result is -
[aah, abh]

If you can use Java 8 then you can build in features to filter your list:
public static void main(String[] args) throws Exception {
final List<String> list = new ArrayList<>();
list.add("cat");
list.add("corn");
list.add("dog");
System.out.println(filter(list, "c"));
}
private static List<String> filter(final Collection<String> source, final String prefix) {
return source.stream().filter(item -> item.startsWith(prefix)).collect(Collectors.toList());
}
This uses the filter method to filter each list item which starts with the String of the prefix argument.
The output is:
[cat, corn]

Comparing an ArrayList of ArrayLists to an ArrayList

In the below example, how would I go about comparing the contents of ArrayList al4 to either of the other ArrayLists? In the same manner I compared al1 to al2.
import java.util.ArrayList;
public class Details {
public static void main(String[] args) {
ArrayList<String> al1 = new ArrayList<String>();
al1.add("hi");
al1.add("How are you");
al1.add("Good Morning");
al1.add("bye");
al1.add("Good night");
ArrayList<String> al2 = new ArrayList<String>();
al2.add("Howdy");
al2.add("Good Evening");
al2.add("bye");
al2.add("Good night");
ArrayList<ArrayList<String>> al4 = new ArrayList<ArrayList<String>>();
al4.add(al1);
// Storing the comparison output in ArrayList<String>
ArrayList<String> al3 = new ArrayList<String>();
for (String temp : al1)
al3.add(al2.contains(temp) ? "Yes" : "No");
System.out.println(al3);
}
}
Output for al3
[No, No, No, Yes, Yes]

It depends a little on what you mean. You can't directly compare al4 with al2, because they're of different types: one is an ArrayList<String> and one is an ArrayList<ArrayList<String>>.
When you compare al1 and al2, what you're doing is determining, for each element of al1, whether it occurs anywhere in al2. I presume this is what you want. You are not deciding whether it occurs in the same place in al2.
The first thing you should do is improve this comparison, which is rather inefficient. Since you don't care about ordering of al2, you should rewrite the comparison like this:
ArrayList<String> al3 = new ArrayList<String>();
Set<String> al2set = new HashSet<String>(al2);
for (String temp : al1)
al3.add(al2set.contains(temp) ? "Yes" : "No");
This is a lot more efficient because you don't have to traverse the whole of al2 for each lookup.
Now, if you want to compare your list of lists with a list, I can only assume that you mean you want to tell, for each element of any of the lists in al4, whether it occurs in al2. If so, then you want
ArrayList<String> al3 = new ArrayList<String>();
Set<String> al2set = new HashSet<String>(al2);
for (List<String> tempList : al4)
for (String temp : tempList)
al3.add(al2set.contains(temp) ? "Yes" : "No");
This will give you one flat ArrayList<String>, recording, for each element of the lists of al4, whether that element was in al2 somewhere.
It is possible that you want your result to be an ArrayList<ArrayList<String>>, so that the result mimics the structure of al4, in which case it's only slightly more complicated:
ArrayList<ArrayList<String>> al3list = new ArrayList<ArrayList<String>>();
Set<String> al2set = new HashSet<String>(al2);
for (List<String> tempList : al4) {
ArrayList<String> al3 = new ArrayList<String>();
for (String temp : tempList)
al3.add(al2set.contains(temp) ? "Yes" : "No");
al3list.add(al3);
}
After executing this, a3list will have the same structure as al4, and each element of each list will be either "Yes" or "No", according to whether the corresponding element in al4 was contained somewhere in al2.

You can create a method like this.
public static boolean[] compare(ArrayList<String> al1, ArrayList<String> al2) {
boolean result[] = new boolean[al1.size()]; // result
int i = 0; // index
/*
* Test if an element of al1 occurs in al2
*/
for(String s : al1) {
if(al2.contains(s)) {
result[i] = true;
}
else {
result[i] = false;
}
i++; // increment
}
return result;
}
And after you can call this method and display the result
boolean result[] = compare(al1, al2);
for(int i = 0; i < result.length; i++) {
System.out.print(result[i] + " ");
}

How to Count Unique Values in an ArrayList?

I have to count the number of unique words from a text document using Java. First I had to get rid of the punctuation in all of the words. I used the Scanner class to scan each word in the document and put in an String ArrayList.
So, the next step is where I'm having the problem! How do I create a method that can count the number of unique Strings in the array?
For example, if the array contains apple, bob, apple, jim, bob; the number of unique values in this array is 3.
public countWords() {
try {
Scanner scan = new Scanner(in);
while (scan.hasNext()) {
String words = scan.next();
if (words.contains(".")) {
words.replace(".", "");
}
if (words.contains("!")) {
words.replace("!", "");
}
if (words.contains(":")) {
words.replace(":", "");
}
if (words.contains(",")) {
words.replace(",", "");
}
if (words.contains("'")) {
words.replace("?", "");
}
if (words.contains("-")) {
words.replace("-", "");
}
if (words.contains("‘")) {
words.replace("‘", "");
}
wordStore.add(words.toLowerCase());
}
} catch (FileNotFoundException e) {
System.out.println("File Not Found");
}
System.out.println("The total number of words is: " + wordStore.size());
}

Are you allowed to use Set? If so, you HashSet may solve your problem. HashSet doesn't accept duplicates.
HashSet noDupSet = new HashSet();
noDupSet.add(yourString);
noDupSet.size();
size() method returns number of unique words.
If you have to really use ArrayList only, then one way to achieve may be,
1) Create a temp ArrayList
2) Iterate original list and retrieve element
3) If tempArrayList doesn't contain element, add element to tempArrayList

Starting from Java 8 you can use Stream:
After you add the elements in your ArrayList:
long n = wordStore.stream().distinct().count();
It converts your ArrayList to a stream and then it counts only the distinct elements.

I would advice to use HashSet. This automatically filters the duplicate when calling add method.

Although I believe a set is the easiest solution, you can still use your original solution and just add an if statement to check if value already exists in the list before you do your add.
if( !wordstore.contains( words.toLowerCase() )
wordStore.add(words.toLowerCase());
Then the number of words in your list is the total number of unique words (ie: wordStore.size() )

This general purpose solution takes advantage of the fact that the Set abstract data type does not allow duplicates. The Set.add() method is specifically useful in that it returns a boolean flag indicating the success of the 'add' operation. A HashMap is used to track the occurrence of each original element. This algorithm can be adapted for variations of this type of problem. This solution produces O(n) performance..
public static void main(String args[])
{
String[] strArray = {"abc", "def", "mno", "xyz", "pqr", "xyz", "def"};
System.out.printf("RAW: %s ; PROCESSED: %s \n",Arrays.toString(strArray), duplicates(strArray).toString());
}
public static HashMap<String, Integer> duplicates(String arr[])
{
HashSet<String> distinctKeySet = new HashSet<String>();
HashMap<String, Integer> keyCountMap = new HashMap<String, Integer>();
for(int i = 0; i < arr.length; i++)
{
if(distinctKeySet.add(arr[i]))
keyCountMap.put(arr[i], 1); // unique value or first occurrence
else
keyCountMap.put(arr[i], (Integer)(keyCountMap.get(arr[i])) + 1);
}
return keyCountMap;
}
RESULTS:
RAW: [abc, def, mno, xyz, pqr, xyz, def] ; PROCESSED: {pqr=1, abc=1, def=2, xyz=2, mno=1}

You can create a HashTable or HashMap as well. Keys would be your input strings and Value would be the number of times that string occurs in your input array. O(N) time and space.
Solution 2:
Sort the input list.
Similar strings would be next to each other.
Compare list(i) to list(i+1) and count the number of duplicates.

In shorthand way you can do it as follows...
ArrayList<String> duplicateList = new ArrayList<String>();
duplicateList.add("one");
duplicateList.add("two");
duplicateList.add("one");
duplicateList.add("three");
System.out.println(duplicateList); // prints [one, two, one, three]
HashSet<String> uniqueSet = new HashSet<String>();
uniqueSet.addAll(duplicateList);
System.out.println(uniqueSet); // prints [two, one, three]
duplicateList.clear();
System.out.println(duplicateList);// prints []
duplicateList.addAll(uniqueSet);
System.out.println(duplicateList);// prints [two, one, three]

public class UniqueinArrayList {
public static void main(String[] args) {
StringBuffer sb=new StringBuffer();
List al=new ArrayList();
al.add("Stack");
al.add("Stack");
al.add("over");
al.add("over");
al.add("flow");
al.add("flow");
System.out.println(al);
Set s=new LinkedHashSet(al);
System.out.println(s);
Iterator itr=s.iterator();
while(itr.hasNext()){
sb.append(itr.next()+" ");
}
System.out.println(sb.toString().trim());
}
}

3 distinct possible solutions:
Use HashSet as suggested above.
Create a temporary ArrayList and store only unique element like below:
public static int getUniqueElement(List<String> data) {
List<String> newList = new ArrayList<>();
for (String eachWord : data)
if (!newList.contains(eachWord))
newList.add(eachWord);
return newList.size();
}
Java 8 solution
long count = data.stream().distinct().count();