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Remove duplicates from an arraylist with strings

I have an arraylist that looks like this:
public static ArrayList<ArrayList<String[]>> x = new ArrayList<>();
I store groups of 2 persons in a pair. For example:
[Person1, Person2]
[Person3, Person4]
The algorithm I use right now still makes duplicates, I've tried out hashmaps and iterating through them with for loop but they just give me back the original list.
This is the code:
package com.company;
import java.io.FileWriter;
import java.io.IOException;
import java.util.*;
public class createGroups
{
public static ArrayList<ArrayList<String[]>> x = new ArrayList<>();
public static void main(String[] args){
//Define names
String[] names = {"Person1", "Person2", "Person3", "Person4"};
try
{
//Create combinations. In a try catch because of the saveFile method.
combination(names, 0, 2);
//Print all the pairs in the Arraylist x
printPairs();
} catch (IOException e)
{
e.printStackTrace();
}
}
static void combination(String[] data, int offset, int group_size) throws IOException
{
if(offset >= data.length)
{
//Create new Arraylist called foo
ArrayList<String[]> foo = new ArrayList<>();
//Create a pair of 2 (data.length = 4 / group_size = 2)
for(int i = 0; i < data.length / group_size; i++)
{
//Add the pair to foo.
foo.add(Arrays.copyOfRange(data, 2 * i, 2 * (i + 1)));
}
//Add foo to x
x.add(foo);
//saveFile(foo);
}
for(int i = offset; i < data.length; i++){
for(int j = i + 1; j < data.length; j++){
swap(data, offset, i);
swap(data, offset + 1, j);
combination(data, offset + group_size, group_size);
swap(data, offset + 1, j);
swap(data, offset, i);
}
}
}
public static void printPairs(){
//Print all pairs
for(ArrayList<String[]> q : x){
for(String[] s : q){
System.out.println(Arrays.toString(s));
}
System.out.println("\n");
}
}
private static void swap(String[] data, int a, int b){
//swap the data around.
String t = data[a];
data[a] = data[b];
data[b] = t;
}
}
The output right now is this:
Output
Every group of 4 names is a 'list' of pairs (Not really a list but that's what I call it)
And this is the desired output:
Desired output
But then you can see that the first and the last list of pairs are basically the same how do I change that in my combination method
The question:
How can I change my combination method so that it doesn't create duplicate groups.
And how can I make the list smaller (The desired output) when printing the created lists.
If I wasn't clear enough or if I didn't explain what I want very well, let me know. I'll try to make it clearer.

Create an object similar to this. It takes 4 strings (2 pairs). Puts the strings into array and sorts this array. That means any combination of strings you put in will be converted into one sorted combination, but the object internaly remembers which person is person1, person2, ...
private class TwoPairs {
private final String person1;
private final String person2;
private final String person3;
private final String person4;
private final String[] persons;
TwoPairs(String person1, String person2, String person3, String person4) {
this.person1 = person1;
this.person2 = person2;
this.person3 = person3;
this.person4 = person4;
persons = new String[4];
persons[0] = person1;
persons[1] = person2;
persons[2] = person3;
persons[3] = person4;
// if we sort array of persons it will convert
// any input combination into single (sorted) combination
Arrays.sort(persons); // sort on 4 objects should be fast
// hashCode and equals will be comparing this sorted array
// and ignore the actual order of inputs
}
// compute hashcode from sorted array
#Override
public int hashCode() {
return Arrays.hashCode(persons);
}
// objects with equal persons arrays are considered equal
#Override
public boolean equals(Object obj) {
if (this == obj) return true;
if (obj == null) return false;
if (getClass() != obj.getClass()) return false;
TwoPairs other = (TwoPairs) obj;
if (!Arrays.equals(persons, other.persons)) return false;
return true;
}
// add methods which you might need
// getters for individual persons
// String getPerson1() { return person1; }
// or perhaps pairs of persons
// String[] getPair1() { return new String[] {person1, person2}; }
// add sensible toString method if you need it
}
Your ArrayList x will change like this
ArrayList<TwoPairs> x = new ArrayList<TwoPairs>();
before adding new TwoPairs object into x check if this list already contains this object.
if (!x.contains(twoPairsObject)) {
x.add(twoPairsObject);
}

Move an element up in the list using Comparator

I have an ArrayList in Java :
{"PatMic", "PatientDoc", "Phram", "Patnet", "PatientA"}
All the elements have a number assigned : PatMic = 20, PatientDoc = 30, Phram = 40, Patnet = 50, PatientA = 60.
And my current Comparator :
Comparator<String> comparator = new Comparator<String>() {
#Override
public int compare(final String o1, final String o2) {
final int numbr1 = getElementNumber(); //Returns element's number in a list
final int numbr2 = getElementNumber();
if (numbr1 > numbr2 ) {
return 1;
} else if (numbr1 < numbr2 ) {
return -1;
}
return 0;
}
};
Collections.sort(strings, comparator);
I do not want to change the assigned numbers to each element but would want to move the element PatientA in between PatMic and PatientDoc so the modified list should look like :
{"PatMic", "PatientA" "PatientDoc", "Phram", "Patnet"}
Could someone please suggest how to achieve this? I tried many ways to modify the existing Comparator logic but in vain. Thank you.

You are trying to sort based on some inherent value associated with a String. Therefore, sorting on a String itself is probably not correct. What you probably want to use is either a custom object (implement equals, hashCode and the interface Comparable), or an enum type. This will allow you to change the internal state of these objects explicitly, which will manifest itself naturally when using a Comparator. For example, using a class:
class MyClass implements Comparable
{
private String name;
private int value;
//Constructor
public MyClass(String s, int v)
{
name = s;
value = v;
}
//Getters and setters
//Implement comparing method
}
Then you can use these objects in place of your Strings:
//...
MyClass patMic = new MyClass("PatMic", 20);
// So on..

First, you should give you comparator sufficient knowledge about what it should do. I mean you should have some data available to comparator that says something like "okay, sort them all by associated number except this one - place it right here". "Right here" could be anything that points exact position, I gonna choose "before that element".
So here we go
public void sortWithException(List<String> data, final Map<String, Integer> numbers, final String element, final String next) {
Collections.sort(data, new Comparator<String>() {
#Override
public int compare(String first, String second) {
if (first.equals(element) || second.equals(element)) { //the exception
Integer nextNumber = numbers.get(next);
Integer firstNumber = numbers.get(first);
Integer secondNumber = numbers.get(second);
if (first.equals(element)) {
if (next == null) // placing the exception after ANY element
return 1;
return secondNumber >= nextNumber ? -1 : 1; //placing the element before next and after all next's predecessors
} else { // second.equals(element)
if (next == null)
return -1;
return firstNumber >= nextNumber ? 1 : -1;
}
} else { //normal sort
return numbers.get(first) - numbers.get(second);
}
}
});
}
and call it like sortWithException(data, numbers, "PatientA", "PatientDoc")
Note that i used Map for associated numbers, you should probably use your own method to get those numbers.

How to get N most often words in given text, sorted from max to min?

I have been given a large text as input. I have made a HashMap that stores each different word as a key, and number of times that occurs as value (Integer).
Now I have to make a method called mostOften(int k):List that return a List that gives the first k-words that from max number of occurrence to min number of occurrence ( descending order ) using the HashMap that I have made before.
The problem is that whenever 2 words have the same number of occurrence, then they should be sorted alphabetically.
The first idea that was on my mind was to swap keys and values of the given HashMap, and put it into TreeMap and TreeMap will sort the words by the key(Integer - number of occurrence of the word ) and then just pop the last/first K-entries from the TreeMap.
But I will have collision for sure, when the number of 2 or 3 words are the same. I will compare the words alphabetically but what Integer should I put as a key of the second word comming.
Any ideas how to implement this, or other options ?

Hints:
Look at the javadocs for the Collections.sort methods ... both of them!
Look at the javadocs for Map.entries().
Think about how to implement a Comparator that compares instances of a class with two fields, using the 2nd as a "tie breaker" when the other compares as equal.

Here's the solution with I come up.
First you create a class MyWord that can store the String value of the word and the number of occurences it appears.
You implement the Comparable interface for this class to sort by occurences first and then alphabetically if the number of occurences is the same
Then for the most often method, you create a new List of MyWord from your original map. You add the entries of this to your List
You sort this list
You take the k-first items of this list using subList
You add those Strings to the List<String> and you return it
public class Test {
public static void main(String [] args){
Map<String, Integer> m = new HashMap<>();
m.put("hello",5);
m.put("halo",5);
m.put("this",2);
m.put("that",2);
m.put("good",1);
System.out.println(mostOften(m, 3));
}
public static List<String> mostOften(Map<String, Integer> m, int k){
List<MyWord> l = new ArrayList<>();
for(Map.Entry<String, Integer> entry : m.entrySet())
l.add(new MyWord(entry.getKey(), entry.getValue()));
Collections.sort(l);
List<String> list = new ArrayList<>();
for(MyWord w : l.subList(0, k))
list.add(w.word);
return list;
}
}
class MyWord implements Comparable<MyWord>{
public String word;
public int occurence;
public MyWord(String word, int occurence) {
super();
this.word = word;
this.occurence = occurence;
}
#Override
public int compareTo(MyWord arg0) {
int cmp = Integer.compare(arg0.occurence,this.occurence);
return cmp != 0 ? cmp : word.compareTo(arg0.word);
}
#Override
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + occurence;
result = prime * result + ((word == null) ? 0 : word.hashCode());
return result;
}
#Override
public boolean equals(Object obj) {
if (this == obj)
return true;
if (obj == null)
return false;
if (getClass() != obj.getClass())
return false;
MyWord other = (MyWord) obj;
if (occurence != other.occurence)
return false;
if (word == null) {
if (other.word != null)
return false;
} else if (!word.equals(other.word))
return false;
return true;
}
}
Output : [halo, hello, that]

In addition to your Map to store word counts I would use a PriorityQueue of fixed size K (with natural order). It will allow to reach O(N) complexity. Here is a code which use this approach:
In constructor we are reading input stream word by word filling the counters in the Map.
In the same time we are updating priority queue keeping it's max size = K (we need count top K words)
public class TopNWordsCounter
{
public static class WordCount
{
String word;
int count;
public WordCount(String word)
{
this.word = word;
this.count = 1;
}
}
private PriorityQueue<WordCount> pq;
private Map<String, WordCount> dict;
public TopNWordsCounter(Scanner scanner)
{
pq = new PriorityQueue<>(10, new Comparator<WordCount>()
{
#Override
public int compare(WordCount o1, WordCount o2)
{
return o2.count-o1.count;
}
});
dict = new HashMap<>();
while (scanner.hasNext())
{
String word = scanner.next();
WordCount wc = dict.get(word);
if (wc == null)
{
wc = new WordCount(word);
dict.put(word, wc);
}
if (pq.contains(wc))
{
pq.remove(wc);
wc.count++;
pq.add(wc);
}
else
{
wc.count++;
if (pq.size() < 10 || wc.count >= pq.peek().count)
{
pq.add(wc);
}
}
if (pq.size() > 10)
{
pq.poll();
}
}
}
public List<String> getTopTenWords()
{
Stack<String> topTen = new Stack<>();
while (!pq.isEmpty())
{
topTen.add(pq.poll().word);
}
return topTen;
}
}

Is it possible to find out if a value exists twice in an arraylist?

I have an integer arraylist..
ArrayList <Integer> portList = new ArrayList();
I need to check if a specific integer has already been entered twice. Is this possible in Java?

You could use something like this to see how many times a specific value is there:
System.out.println(Collections.frequency(portList, 1));
// There can be whatever Integer, and I use 1, so you can understand
And to check if a specific value is there more than once you could use something like this:
if ( (Collections.frequency(portList, x)) > 1 ){
System.out.println(x + " is in portList more than once ");
}

My solution
public static boolean moreThanOnce(ArrayList<Integer> list, int searched)
{
int numCount = 0;
for (int thisNum : list) {
if (thisNum == searched)
numCount++;
}
return numCount > 1;
}

If you are looking to do this in one method, then no. However, you could do it in two steps if you need to simply find out if it exists at least more than once in the List. You could do
int first = list.indexOf(object)
int second = list.lastIndexOf(object)
// Don't forget to also check to see if either are -1, the value does not exist at all.
if (first == second) {
// No Duplicates of object appear in the list
} else {
// Duplicate exists
}

This will tell you if you have at least two same values in your ArrayList:
int first = portList.indexOf(someIntValue);
int last = portList.lastIndexOf(someIntValue);
if (first != -1 && first != last) {
// someIntValue exists more than once in the list (not sure how many times though)
}
If you really want to know how many duplicates of a given value you have, you need to iterate through the entire array. Something like this:
/**
* Will return a list of all indexes where the given value
* exists in the given array. The list will be empty if the
* given value does not exist at all.
*
* #param List<E> list
* #param E value
* #return List<Integer> a list of indexes in the list
*/
public <E> List<Integer> collectFrequency(List<E> list, E value) {
ArrayList<Integer> freqIndex = new ArrayList<Integer>();
E item;
for (int i=0, len=list.size(); i<len; i++) {
item = list.get(i);
if ((item == value) || (null != item && item.equals(value))) {
freqIndex.add(i);
}
}
return freqIndex;
}
if (!collectFrequency(portList, someIntValue).size() > 1) {
// Duplicate value
}
Or using the already availble method:
if (Collections.frequency(portList, someIntValue) > 1) {
// Duplicate value
}

Set portSet = new HashSet<Integer>();
portSet.addAll(portList);
boolean listContainsDuplicates = portSet.size() != portList.size();

I used the following solution to find out whether an ArrayList contains a number more than once. This solution comes very close to the one listed by user3690146, but it does not use a helper variable at all. After running it, you get "The number is listed more than once" as a return message.
public class Application {
public static void main(String[] args) {
ArrayList<Integer> list = new ArrayList<Integer>();
list.add(4);
list.add(8);
list.add(1);
list.add(8);
int number = 8;
if (NumberMoreThenOnceInArray(list, number)) {
System.out.println("The number is listed more than once");
} else {
System.out.println("The number is not listed more than once");
}
}
public static boolean NumberMoreThenOnceInArray(ArrayList<Integer> list, int whichNumber) {
int numberCounter = 0;
for (int number : list) {
if (number == whichNumber) {
numberCounter++;
}
}
if (numberCounter > 1) {
return true;
}
return false;
}
}

Here is my solution (in Kotlin):
// getItemsMoreThan(list, 2) -> [4.45, 333.45, 1.1, 4.45, 333.45, 2.05, 4.45, 333.45, 2.05, 4.45] -> {4.45=4, 333.45=3}
// getItemsMoreThan(list, 1)-> [4.45, 333.45, 1.1, 4.45, 333.45, 2.05, 4.45, 333.45, 2.05, 4.45] -> {4.45=4, 333.45=3, 2.05=2}
fun getItemsMoreThan(list: List<Any>, moreThan: Int): Map<Any, Int> {
val mapNumbersByElement: Map<Any, Int> = getHowOftenItemsInList(list)
val findItem = mapNumbersByElement.filter { it.value > moreThan }
return findItem
}
// Return(map) how often an items is list.
// E.g.: [16.44, 200.00, 200.00, 33.33, 200.00, 0.00] -> {16.44=1, 200.00=3, 33.33=1, 0.00=1}
fun getHowOftenItemsInList(list: List<Any>): Map<Any, Int> {
val mapNumbersByItem = list.groupingBy { it }.eachCount()
return mapNumbersByItem
}

By looking at the question, we need to find out whether a value exists twice in an ArrayList. So I believe that we can reduce the overhead of "going through the entire list just to check whether the value only exists twice" by doing the simple check below.
public boolean moreThanOneMatch(int number, ArrayList<Integer> list) {
int count = 0;
for (int num : list) {
if (num == number) {
count ++ ;
if (count == 2) {
return true;
}
}
}
return false;
}

Using a method to sort a Singly Linked List in Java

Implementing a Linked List, store up to 10 names, ordered in first in First Out. Then implement two methods, one of which to sort it alphabetically by last names. This is where I am having some trouble. Here is what I tried:
Recursion. The method calls two nodes, compare them, swap if needed and then calls itself. Doesn't work with odd number of names and tends to be full bugs.
Collection but I don't know enough about it to use it effectively.
Sorting algorithms (ex. bubble sort): I can go though the list but have a hard time getting the nodes to swap.
My question is: What is the easiest way to do this?
public class List
{
public class Link
{
public String firstName;
public String middleName;
public String lastName;
public Link next = null;
Link(String f, String m, String l)
{
firstName = f;
middleName = m;
lastName = l;
}
}
private Link first_;
private Link last_;
List()
{
first_ = null;
last_ = null;
}
public boolean isEmpty()
{
return first_ == null;
}
public void insertFront(String f, String m, String l)
{
Link name = new Link(f, m, l);
if (first_ == null)
{
first_ = name;
last_ = name;
}
else
{
last_.next = name;
last_ = last_.next;
}
}
public String removeFront()
{
String f = first_.firstName;
String m = first_.middleName;
String l = first_.lastName;
first_ = first_.next;
return f + " " + m + " " + l;
}
public String findMiddle(String f, String l)
{
Link current = first_;
while (current != null && current.firstName.compareTo(f) != 0 && current.lastName.compareTo(l) != 0)
{
current = current.next;
}
if (current == null)
{
return "Not in list";
}
return "That person's middle name is " + current.middleName;
}
}
public class NamesOfFriends
{
public static void main(String[] args)
{
List listOfnames = new List();
Scanner in = new Scanner(System.in);
for(int i = 0; i < 3; i++)
{
if(i == 0)
{
System.out.println("Please enter the first, middle and last name?");
listOfnames.insertFront(in.next(), in.next(),in.next());
}
else
{
System.out.println("Please enter the next first, middle and last name");
listOfnames.insertFront(in.next(), in.next(),in.next());
}
}
System.out.println("To find the middle name, please enter the first and last name of the person.");
System.out.println(listOfnames.findMiddle(in.next(),in.next()));
}
}
Edit
After working on it a bit, I figured out how to go about sorting it. For that purpose, I am trying to implement a remove method that can remove a node anywhere on the list. While it does compile, it doesn't do anything when I run the program.
public Link remove(String lastName)
{
Link current_ = first_;
Link prior_ = null;
Link temp_ = null;
while (current_ != null && current_.lastName.compareTo(lastName) != 0)
{
prior_ = current_;
current_ = current_.next;
}
if (current_ != null)
{
if (current_ == last_)
{
temp_ = last_;
last_ = prior_;
}
else if (prior_ == null)
{
temp_ = first_;
first_ = first_.next;
}
}
else
{
temp_ = current_;
prior_.next = current_.next;
}
return temp_;
}

2: Collections is the easiest, but it seems to be not allowed in your homework
3: BubbleSort is easy but the worst known sorting algo, however for your homework it probably is ok
1: This is the same as bubble sort, but is prefered to be done without recursion
In BubbleSort you loop through your elements again and again till no swaps are neeeded anymore, then you are ready.

Collection is the easiest way to accomplish this.
Implement Comparable
Override hashcode and equals
Collection.sort()

You already has the linked list implemented, that is good.
Have you considered implementing MergeSort as the sorting algorithm? Being the divide&conquer algorithm, you will always end up with only two elements to form a list with.
The merge part is going to be trickier, but also easy. Basically you just create a new list and start filling it up with elements you get by comparing the first values of the two merging sets.
So for instance if you have two sets to merge:
[A]->[C]->[D]
[B]->[E]->[F]
the mergin process will go:
[A]
[C]->[D]
[B]->[E]->[F]
[A]->[B]
[C]->[D]
[E]->[F]
[A]->[B]->[C]
[D]
[E]->[F]
[A]->[B]->[C]->[D]
[E]->[F]
[A]->[B]->[C]->[D]->[E]
[F]
[A]->[B]->[C]->[D]->[E]->[F]