Related
I'm trying to answer this question:
Program the method findIngredients. This method takes in a String called
foodInStock, and an ArrayList of Strings called ingredients. The method should return an
ArrayList of ingredients that were not found in foodInStock.
for example if:
foodInStock = “tomatopotatocornturkeycarrotstuffing”
ingredients = {“potato”, “corn”, “salt”, “chicken”, “turkey”}
returns {“salt”, “chicken”}
I tried writing some code but for some reason everything is getting removed when I use the above example on my program. Where did my program go wrong?
Here's my code:
public static ArrayList<String> findIngredients(String foodInStock, ArrayList<String> ingredients){
ArrayList<String> ingredientsNotFound = new ArrayList<String>();
int i = 0;
for (; i < ingredients.size(); i++) {
for (int x = 0; x < foodInStock.length()-(ingredients.get(i).length())+1; x++) {
if (ingredients.get(i) == foodInStock.substring(x, (x + ingredients.get(i).length()))) {
ingredients.remove(i);
i = 0;
break;
}
}
}
ingredients = ingredientsNotFound;
return ingredientsNotFound;
}
I think there are two main things to cover here.
First, the way to build the final result. You are currently removing items from the original input; a better strategy is to add items to a new list (partially because it's simpler to think about and partially because you generally don't want to modify a list while iterating over it).
You also are, probably accidentally, overwriting your list with an empty list at the end.
Second, the way to determine whether or not the ingredient is in the string input. Rather than looping over the whole string and inspecting substrings, you can instead use the indexOf() method to see whether or not the string includes the current item.
public static ArrayList<String> findIngredients(String foodInStock, ArrayList<String> ingredients) {
ArrayList<String> results = new ArrayList<>();
for (String ingredient : ingredients) {
if (foodInStock.indexOf(ingredient) == -1) {
results.add(ingredient);
}
}
return results;
}
Here we initialize a new list for the results. We then loop over every individual ingredient in the input list, and ask whether or not that ingredient is present in the string input. When it is not (indexOf() returns -1), we add it to the results list. At the end, the results contains every ingredient not found.
How could I go about detecting (returning true/false) whether an ArrayList contains more than one of the same element in Java?
Many thanks,
Terry
Edit
Forgot to mention that I am not looking to compare "Blocks" with each other but their integer values. Each "block" has an int and this is what makes them different.
I find the int of a particular Block by calling a method named "getNum" (e.g. table1[0][2].getNum();
Simplest: dump the whole collection into a Set (using the Set(Collection) constructor or Set.addAll), then see if the Set has the same size as the ArrayList.
List<Integer> list = ...;
Set<Integer> set = new HashSet<Integer>(list);
if(set.size() < list.size()){
/* There are duplicates */
}
Update: If I'm understanding your question correctly, you have a 2d array of Block, as in
Block table[][];
and you want to detect if any row of them has duplicates?
In that case, I could do the following, assuming that Block implements "equals" and "hashCode" correctly:
for (Block[] row : table) {
Set set = new HashSet<Block>();
for (Block cell : row) {
set.add(cell);
}
if (set.size() < 6) { //has duplicate
}
}
I'm not 100% sure of that for syntax, so it might be safer to write it as
for (int i = 0; i < 6; i++) {
Set set = new HashSet<Block>();
for (int j = 0; j < 6; j++)
set.add(table[i][j]);
...
Set.add returns a boolean false if the item being added is already in the set, so you could even short circuit and bale out on any add that returns false if all you want to know is whether there are any duplicates.
Improved code, using return value of Set#add instead of comparing the size of list and set.
public static <T> boolean hasDuplicate(Iterable<T> all) {
Set<T> set = new HashSet<T>();
// Set#add returns false if the set does not change, which
// indicates that a duplicate element has been added.
for (T each: all) if (!set.add(each)) return true;
return false;
}
With Java 8+ you can use Stream API:
boolean areAllDistinct(List<Block> blocksList) {
return blocksList.stream().map(Block::getNum).distinct().count() == blockList.size();
}
If you are looking to avoid having duplicates at all, then you should just cut out the middle process of detecting duplicates and use a Set.
Improved code to return the duplicate elements
Can find duplicates in a Collection
return the set of duplicates
Unique Elements can be obtained from the Set
public static <T> List getDuplicate(Collection<T> list) {
final List<T> duplicatedObjects = new ArrayList<T>();
Set<T> set = new HashSet<T>() {
#Override
public boolean add(T e) {
if (contains(e)) {
duplicatedObjects.add(e);
}
return super.add(e);
}
};
for (T t : list) {
set.add(t);
}
return duplicatedObjects;
}
public static <T> boolean hasDuplicate(Collection<T> list) {
if (getDuplicate(list).isEmpty())
return false;
return true;
}
I needed to do a similar operation for a Stream, but couldn't find a good example. Here's what I came up with.
public static <T> boolean areUnique(final Stream<T> stream) {
final Set<T> seen = new HashSet<>();
return stream.allMatch(seen::add);
}
This has the advantage of short-circuiting when duplicates are found early rather than having to process the whole stream and isn't much more complicated than just putting everything in a Set and checking the size. So this case would roughly be:
List<T> list = ...
boolean allDistinct = areUnique(list.stream());
If your elements are somehow Comparable (the fact that the order has any real meaning is indifferent -- it just needs to be consistent with your definition of equality), the fastest duplicate removal solution is going to sort the list ( 0(n log(n)) ) then to do a single pass and look for repeated elements (that is, equal elements that follow each other) (this is O(n)).
The overall complexity is going to be O(n log(n)), which is roughly the same as what you would get with a Set (n times long(n)), but with a much smaller constant. This is because the constant in sort/dedup results from the cost of comparing elements, whereas the cost from the set is most likely to result from a hash computation, plus one (possibly several) hash comparisons. If you are using a hash-based Set implementation, that is, because a Tree based is going to give you a O( n log²(n) ), which is even worse.
As I understand it, however, you do not need to remove duplicates, but merely test for their existence. So you should hand-code a merge or heap sort algorithm on your array, that simply exits returning true (i.e. "there is a dup") if your comparator returns 0, and otherwise completes the sort, and traverse the sorted array testing for repeats. In a merge or heap sort, indeed, when the sort is completed, you will have compared every duplicate pair unless both elements were already in their final positions (which is unlikely). Thus, a tweaked sort algorithm should yield a huge performance improvement (I would have to prove that, but I guess the tweaked algorithm should be in the O(log(n)) on uniformly random data)
If you want the set of duplicate values:
import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Set;
public class FindDuplicateInArrayList {
public static void main(String[] args) {
Set<String> uniqueSet = new HashSet<String>();
List<String> dupesList = new ArrayList<String>();
for (String a : args) {
if (uniqueSet.contains(a))
dupesList.add(a);
else
uniqueSet.add(a);
}
System.out.println(uniqueSet.size() + " distinct words: " + uniqueSet);
System.out.println(dupesList.size() + " dupesList words: " + dupesList);
}
}
And probably also think about trimming values or using lowercase ... depending on your case.
Simply put:
1) make sure all items are comparable
2) sort the array
2) iterate over the array and find duplicates
To know the Duplicates in a List use the following code:It will give you the set which contains duplicates.
public Set<?> findDuplicatesInList(List<?> beanList) {
System.out.println("findDuplicatesInList::"+beanList);
Set<Object> duplicateRowSet=null;
duplicateRowSet=new LinkedHashSet<Object>();
for(int i=0;i<beanList.size();i++){
Object superString=beanList.get(i);
System.out.println("findDuplicatesInList::superString::"+superString);
for(int j=0;j<beanList.size();j++){
if(i!=j){
Object subString=beanList.get(j);
System.out.println("findDuplicatesInList::subString::"+subString);
if(superString.equals(subString)){
duplicateRowSet.add(beanList.get(j));
}
}
}
}
System.out.println("findDuplicatesInList::duplicationSet::"+duplicateRowSet);
return duplicateRowSet;
}
best way to handle this issue is to use a HashSet :
ArrayList<String> listGroupCode = new ArrayList<>();
listGroupCode.add("A");
listGroupCode.add("A");
listGroupCode.add("B");
listGroupCode.add("C");
HashSet<String> set = new HashSet<>(listGroupCode);
ArrayList<String> result = new ArrayList<>(set);
Just print result arraylist and see the result without duplicates :)
This answer is wrriten in Kotlin, but can easily be translated to Java.
If your arraylist's size is within a fixed small range, then this is a great solution.
var duplicateDetected = false
if(arrList.size > 1){
for(i in 0 until arrList.size){
for(j in 0 until arrList.size){
if(i != j && arrList.get(i) == arrList.get(j)){
duplicateDetected = true
}
}
}
}
private boolean isDuplicate() {
for (int i = 0; i < arrayList.size(); i++) {
for (int j = i + 1; j < arrayList.size(); j++) {
if (arrayList.get(i).getName().trim().equalsIgnoreCase(arrayList.get(j).getName().trim())) {
return true;
}
}
}
return false;
}
String tempVal = null;
for (int i = 0; i < l.size(); i++) {
tempVal = l.get(i); //take the ith object out of list
while (l.contains(tempVal)) {
l.remove(tempVal); //remove all matching entries
}
l.add(tempVal); //at last add one entry
}
Note: this will have major performance hit though as items are removed from start of the list.
To address this, we have two options. 1) iterate in reverse order and remove elements. 2) Use LinkedList instead of ArrayList. Due to biased questions asked in interviews to remove duplicates from List without using any other collection, above example is the answer. In real world though, if I have to achieve this, I will put elements from List to Set, simple!
/**
* Method to detect presence of duplicates in a generic list.
* Depends on the equals method of the concrete type. make sure to override it as required.
*/
public static <T> boolean hasDuplicates(List<T> list){
int count = list.size();
T t1,t2;
for(int i=0;i<count;i++){
t1 = list.get(i);
for(int j=i+1;j<count;j++){
t2 = list.get(j);
if(t2.equals(t1)){
return true;
}
}
}
return false;
}
An example of a concrete class that has overridden equals() :
public class Reminder{
private long id;
private int hour;
private int minute;
public Reminder(long id, int hour, int minute){
this.id = id;
this.hour = hour;
this.minute = minute;
}
#Override
public boolean equals(Object other){
if(other == null) return false;
if(this.getClass() != other.getClass()) return false;
Reminder otherReminder = (Reminder) other;
if(this.hour != otherReminder.hour) return false;
if(this.minute != otherReminder.minute) return false;
return true;
}
}
ArrayList<String> withDuplicates = new ArrayList<>();
withDuplicates.add("1");
withDuplicates.add("2");
withDuplicates.add("1");
withDuplicates.add("3");
HashSet<String> set = new HashSet<>(withDuplicates);
ArrayList<String> withoutDupicates = new ArrayList<>(set);
ArrayList<String> duplicates = new ArrayList<String>();
Iterator<String> dupIter = withDuplicates.iterator();
while(dupIter.hasNext())
{
String dupWord = dupIter.next();
if(withDuplicates.contains(dupWord))
{
duplicates.add(dupWord);
}else{
withoutDupicates.add(dupWord);
}
}
System.out.println(duplicates);
System.out.println(withoutDupicates);
A simple solution for learners.
//Method to find the duplicates.
public static List<Integer> findDublicate(List<Integer> numList){
List<Integer> dupLst = new ArrayList<Integer>();
//Compare one number against all the other number except the self.
for(int i =0;i<numList.size();i++) {
for(int j=0 ; j<numList.size();j++) {
if(i!=j && numList.get(i)==numList.get(j)) {
boolean isNumExist = false;
//The below for loop is used for avoid the duplicate again in the result list
for(Integer aNum: dupLst) {
if(aNum==numList.get(i)) {
isNumExist = true;
break;
}
}
if(!isNumExist) {
dupLst.add(numList.get(i));
}
}
}
}
return dupLst;
}
How to delete the content of an array of objects. If there is other ways to delete a content of an array of objects , please do share.
import java.util.Arrays;
import java.util.Scanner;
import org.apache.commons.lang3.ArrayUtils;
public class Testing {
public static void deleteItem(ItemTracker[] listItems) {
System.out.println("Which item you want to delete? ");
for(int i=0; i < listItems.length; i++) {
if(input.equalsIgnoreCase("Quantity")) {
// Some Code
} else if(input.equalsIgnoreCase("Something"){
ArrayUtils.remove(listItems, i); // This is the part where it should delete .. but it doesnt delete.
}
break;
}
}
}
Change this
ArrayUtils.remove(listItems, i);
to
listItems = ArrayUtils.remove(listItems, i);
As you can see in the JavaDoc, the method does not change the argument listItems, rather it returns a new array with the remaining elements.
Edit
You also need to change your deletion method to
public static ItemTracker[] deleteItem(ItemTracker[] listItems) {
//..
}
So you could return the new array with the remaining elements.
Store the resulting array.
It won't change the original array object.
listItems = ArrayUtils.remove(listItems, i);
Edit: But for using this method you need the change to return type of your method
public static ItemTracker[] deleteItem(ItemTracker[] listItems){
System.out.println("Which item you want to delete? ");
for(int i=0; i < listItems.length; i++) {
if(input.equalsIgnoreCase("Quantity")) {
// Some Code
} else if(input.equalsIgnoreCase("Something"){
listItems = ArrayUtils.remove(listItems, i); // This is the part where it should delete .. but it doesnt delete.
}
break;
}
return listItems;
}
In your case usage of ArrayUtils is incorrect and redundant. You can delete element in next way:
// ...
listItems[i] = null;
// array will looks like [o1, o2, null, o3, o4, ...]
// ...
There is no other way without changing method's return type
Without additional libraries, with temporary list:
Element arrayToRemoveFrom[];
Element toRemove; // should be known already
ArrayList<Element> tmpList = new ArrayList<Element>(Arrays.asList(arrayToRemoveFrom));
tmpList.remove(toRemove);
// any other code processing and removing elements
arrayToRemoveFrom = tmpList.toArray(new Arrays[tmlList.size()]);
ArrayUtils.remove
This method returns a new array with the same elements of the input
array except the element on the specified position. The component type
of the returned array is always the same as that of the input array.
So,you should use it like this
listItems = ArrayUtils.remove(listItems, i);
NOTE
Here we have assign the returned array to current listItem.
As this method does not change the actual array but returns the changed array after removal same as #replace method works for String.
YES. I agree with zvdh I have missed the purpose of your method
because I was more concentrated on removal of element.Sorry for that!!
as this will not actually change the listItem and you need to return the new array which contains the change.
class Arrays
{
public static void main(String[] args)
{
double[] numbers = {6.0, 4.4, 1.9, 2.9, 3.4, 3.5};
java.util.Arrays.sort(numbers);
System.out.print("Ascending order= ");
for (int i = 0; i < numbers.length; i++)
System.out.print(numbers[i] + " ");
System.out.println();
System.out.print("Decending order= ");
for (int i = numbers.length -1; i >= 0; i--)
System.out.print(numbers[i] + " ");
}
}
This solution only displays in reverse order, but it can be changed to reorder the array using the same loop.
I am trying to write a method that takes an ArrayList of Strings as a parameter and that places a string of four asterisks in front of every string of length 4.
However, in my code, I am getting an error in the way I constructed my method.
Here is my mark length class
import java.util.ArrayList;
public class Marklength {
void marklength4(ArrayList <String> themarklength){
for(String n : themarklength){
if(n.length() ==4){
themarklength.add("****");
}
}
System.out.println(themarklength);
}
}
And the following is my main class:
import java.util.ArrayList;
public class MarklengthTestDrive {
public static void main(String[] args){
ArrayList <String> words = new ArrayList<String>();
words.add("Kane");
words.add("Cane");
words.add("Fame");
words.add("Dame");
words.add("Lame");
words.add("Same");
Marklength ish = new Marklength();
ish.marklength4(words);
}
}
Essentially in this case, it should run so it adds an arraylist with a string of "****" placed before every previous element of the array list because the lengths of the strings are all 4.
BTW
This consists of adding another element
I am not sure where I went wrong. Possibly in my for loop?
I got the following error:
Exception in thread "main" java.util.ConcurrentModificationException
at java.util.AbstractList$Itr.checkForComodification(AbstractList.java:372)
at java.util.AbstractList$Itr.next(AbstractList.java:343)
at Marklength.marklength4(Marklength.java:7)
at MarklengthTestDrive.main(MarklengthTestDrive.java:18)
Thank you very much. Help is appreciated.
Let's think about this piece of code, and pretend like you don't get that exception:
import java.util.ArrayList;
public class Marklength {
void marklength4(ArrayList <String> themarklength){
for(String n : themarklength){
if(n.length() ==4){
themarklength.add("****");
}
}
System.out.println(themarklength);
}
}
Ok, so what happens if your list just contains item.
You hit the line if(n.length() ==4){, which is true because you are looking at item, so you go execute its block.
Next you hit the line themarklength.add("****");. Your list now has the element **** at the end of it.
The loop continues, and you get the next item in the list, which happens to be the one you just added, ****.
The next line you hit is if(n.length() ==4){. This is true, so you execute its block.
You go to the line themarklength.add("****");, and add **** to the end of the list.
Do we see a bad pattern here? Yes, yes we do.
The Java runtime environment also knows that this is bad, which is why it prevents something called Concurrent Modification. In your case, this means you cannot modify a list while you are iterating over it, which is what that for loop does.
My best guess as to what you are trying to do is something like this:
import java.util.ArrayList;
public class Marklength {
ArrayList<String> marklength4(ArrayList <String> themarklength){
ArrayList<String> markedStrings = new ArrayList<String>(themarklength.size());
for(String n : themarklength){
if(n.length() ==4){
markedStrings.add("****");
}
markedStrings.add(n);
}
System.out.println(themarklength);
return markedStrings;
}
}
And then:
import java.util.ArrayList;
public class MarklengthTestDrive {
public static void main(String[] args){
ArrayList <String> words = new ArrayList<String>();
words.add("Kane");
words.add("Cane");
words.add("Fame");
words.add("Dame");
words.add("Lame");
words.add("Same");
Marklength ish = new Marklength();
words = ish.marklength4(words);
}
}
This...
if(n.length() ==4){
themarklength.add("****");
}
Is simply trying to add "****" to the end of the list. This fails because the Iterator used by the for-each loop won't allow changes to occur to the underlying List while it's been iterated.
You could create a copy of the List first...
List<String> values = new ArrayList<String>(themarklength);
Or convert it to an array of String
String[] values = themarklength.toArray(new String[themarklength.size()]);
And uses these as you iteration points...
for (String value : values) {
Next, you need to be able to insert a new element into the ArrayList at a specific point. To do this, you will need to know the original index of the value you are working with...
if (value.length() == 4) {
int index = themarklength.indexOf(value);
And then add a new value at the required location...
themarklength.add(index, "****");
This will add the "****" at the index point, pushing all the other entries down
Updated
As has, correctly, been pointed out to me, the use of themarklength.indexOf(value) won't take into account the use case where the themarklength list contains two elements of the same value, which would return the wrong index.
I also wasn't focusing on performance as a major requirement for the providing a possible solution.
Updated...
As pointed out by JohnGarnder and AnthonyAccioly, you could use for-loop instead of a for-each which would allow you to dispense with the themarklength.indexOf(value)
This will remove the risk of duplicate values messing up the index location and improve the overall performance, as you don't need to create a second iterator...
// This assumes you're using the ArrayList as the copy...
for (int index = 0; index < themarklength.size(); index++) {
String value = themarklength.get(index);
if (value.length() == 4) {
themarklength.add(index, "****");
index++;
But which you use is up to you...
The problem is that in your method, you didn't modify each string in the arraylist, but only adds 4 stars to the list. So the correct way to do this is, you need to modify each element of the arraylist and replace the old string with the new one:
void marklength4(ArrayList<String> themarklength){
int index = 0;
for(String n : themarklength){
if(n.length() ==4){
n = "****" + n;
}
themarklength.set(index++, n);
}
System.out.println(themarklength);
}
If this is not what you want but you want to add a new string "**" before each element in the arraylist, then you can use listIterator method in the ArrayList to add new additional element before EACH string if the length is 4.
ListIterator<String> it = themarklength.listIterator();
while(it.hasNext()) {
String name = it.next();
if(name.length() == 4) {
it.previous();
it.add("****");
it.next();
}
}
The difference is: ListIterator allows you to modify the list when iterating through it and also allows you to go backward in the list.
I would use a ListIterator instead of a for each, listiterator.add likely do exactly what you want.
public void marklength4(List<String> themarklength){
final ListIterator<String> lit =
themarklength.listIterator(themarklength.size());
boolean shouldInsert = false;
while(lit.hasPrevious()) {
if (shouldInsert) {
lit.add("****");
lit.previous();
shouldInsert = false;
}
final String n = lit.previous();
shouldInsert = (n.length() == 4);
}
if (shouldInsert) {
lit.add("****");
}
}
Working example
Oh I remember this lovely error from the good old days. The problem is that your ArrayList isn't completely populated by the time the array element is to be accessed. Think of it, you create the object and then immediately start looping it. The object hence, has to populate itself with the values as the loop is going to be running.
The simple way to solve this is to pre-populate your ArrayList.
public class MarklengthTestDrive {
public static void main(String[] args){
ArrayList <String> words = new ArrayList<String>() {{
words.add("Kane");
words.add("Cane");
words.add("Fame");
words.add("Dame");
words.add("Lame");
words.add("Same");
}};
}
}
Do tell me if that fixes it. You can also use a static initializer.
make temporary arraylist, modify this list and copy its content at the end to the original list
import java.util.ArrayList;
public class MarkLength {
void marklength4(ArrayList <String> themarklength){
ArrayList<String> temp = new ArrayList<String>();
for(String n : themarklength){
if(n.length() ==4){
temp.add(n);
temp.add("****");
}
}
themarklength.clear();
themarklength.addAll(temp);
System.out.println(themarklength);
}
}
How could I go about detecting (returning true/false) whether an ArrayList contains more than one of the same element in Java?
Many thanks,
Terry
Edit
Forgot to mention that I am not looking to compare "Blocks" with each other but their integer values. Each "block" has an int and this is what makes them different.
I find the int of a particular Block by calling a method named "getNum" (e.g. table1[0][2].getNum();
Simplest: dump the whole collection into a Set (using the Set(Collection) constructor or Set.addAll), then see if the Set has the same size as the ArrayList.
List<Integer> list = ...;
Set<Integer> set = new HashSet<Integer>(list);
if(set.size() < list.size()){
/* There are duplicates */
}
Update: If I'm understanding your question correctly, you have a 2d array of Block, as in
Block table[][];
and you want to detect if any row of them has duplicates?
In that case, I could do the following, assuming that Block implements "equals" and "hashCode" correctly:
for (Block[] row : table) {
Set set = new HashSet<Block>();
for (Block cell : row) {
set.add(cell);
}
if (set.size() < 6) { //has duplicate
}
}
I'm not 100% sure of that for syntax, so it might be safer to write it as
for (int i = 0; i < 6; i++) {
Set set = new HashSet<Block>();
for (int j = 0; j < 6; j++)
set.add(table[i][j]);
...
Set.add returns a boolean false if the item being added is already in the set, so you could even short circuit and bale out on any add that returns false if all you want to know is whether there are any duplicates.
Improved code, using return value of Set#add instead of comparing the size of list and set.
public static <T> boolean hasDuplicate(Iterable<T> all) {
Set<T> set = new HashSet<T>();
// Set#add returns false if the set does not change, which
// indicates that a duplicate element has been added.
for (T each: all) if (!set.add(each)) return true;
return false;
}
With Java 8+ you can use Stream API:
boolean areAllDistinct(List<Block> blocksList) {
return blocksList.stream().map(Block::getNum).distinct().count() == blockList.size();
}
If you are looking to avoid having duplicates at all, then you should just cut out the middle process of detecting duplicates and use a Set.
Improved code to return the duplicate elements
Can find duplicates in a Collection
return the set of duplicates
Unique Elements can be obtained from the Set
public static <T> List getDuplicate(Collection<T> list) {
final List<T> duplicatedObjects = new ArrayList<T>();
Set<T> set = new HashSet<T>() {
#Override
public boolean add(T e) {
if (contains(e)) {
duplicatedObjects.add(e);
}
return super.add(e);
}
};
for (T t : list) {
set.add(t);
}
return duplicatedObjects;
}
public static <T> boolean hasDuplicate(Collection<T> list) {
if (getDuplicate(list).isEmpty())
return false;
return true;
}
I needed to do a similar operation for a Stream, but couldn't find a good example. Here's what I came up with.
public static <T> boolean areUnique(final Stream<T> stream) {
final Set<T> seen = new HashSet<>();
return stream.allMatch(seen::add);
}
This has the advantage of short-circuiting when duplicates are found early rather than having to process the whole stream and isn't much more complicated than just putting everything in a Set and checking the size. So this case would roughly be:
List<T> list = ...
boolean allDistinct = areUnique(list.stream());
If your elements are somehow Comparable (the fact that the order has any real meaning is indifferent -- it just needs to be consistent with your definition of equality), the fastest duplicate removal solution is going to sort the list ( 0(n log(n)) ) then to do a single pass and look for repeated elements (that is, equal elements that follow each other) (this is O(n)).
The overall complexity is going to be O(n log(n)), which is roughly the same as what you would get with a Set (n times long(n)), but with a much smaller constant. This is because the constant in sort/dedup results from the cost of comparing elements, whereas the cost from the set is most likely to result from a hash computation, plus one (possibly several) hash comparisons. If you are using a hash-based Set implementation, that is, because a Tree based is going to give you a O( n log²(n) ), which is even worse.
As I understand it, however, you do not need to remove duplicates, but merely test for their existence. So you should hand-code a merge or heap sort algorithm on your array, that simply exits returning true (i.e. "there is a dup") if your comparator returns 0, and otherwise completes the sort, and traverse the sorted array testing for repeats. In a merge or heap sort, indeed, when the sort is completed, you will have compared every duplicate pair unless both elements were already in their final positions (which is unlikely). Thus, a tweaked sort algorithm should yield a huge performance improvement (I would have to prove that, but I guess the tweaked algorithm should be in the O(log(n)) on uniformly random data)
If you want the set of duplicate values:
import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Set;
public class FindDuplicateInArrayList {
public static void main(String[] args) {
Set<String> uniqueSet = new HashSet<String>();
List<String> dupesList = new ArrayList<String>();
for (String a : args) {
if (uniqueSet.contains(a))
dupesList.add(a);
else
uniqueSet.add(a);
}
System.out.println(uniqueSet.size() + " distinct words: " + uniqueSet);
System.out.println(dupesList.size() + " dupesList words: " + dupesList);
}
}
And probably also think about trimming values or using lowercase ... depending on your case.
Simply put:
1) make sure all items are comparable
2) sort the array
2) iterate over the array and find duplicates
To know the Duplicates in a List use the following code:It will give you the set which contains duplicates.
public Set<?> findDuplicatesInList(List<?> beanList) {
System.out.println("findDuplicatesInList::"+beanList);
Set<Object> duplicateRowSet=null;
duplicateRowSet=new LinkedHashSet<Object>();
for(int i=0;i<beanList.size();i++){
Object superString=beanList.get(i);
System.out.println("findDuplicatesInList::superString::"+superString);
for(int j=0;j<beanList.size();j++){
if(i!=j){
Object subString=beanList.get(j);
System.out.println("findDuplicatesInList::subString::"+subString);
if(superString.equals(subString)){
duplicateRowSet.add(beanList.get(j));
}
}
}
}
System.out.println("findDuplicatesInList::duplicationSet::"+duplicateRowSet);
return duplicateRowSet;
}
best way to handle this issue is to use a HashSet :
ArrayList<String> listGroupCode = new ArrayList<>();
listGroupCode.add("A");
listGroupCode.add("A");
listGroupCode.add("B");
listGroupCode.add("C");
HashSet<String> set = new HashSet<>(listGroupCode);
ArrayList<String> result = new ArrayList<>(set);
Just print result arraylist and see the result without duplicates :)
This answer is wrriten in Kotlin, but can easily be translated to Java.
If your arraylist's size is within a fixed small range, then this is a great solution.
var duplicateDetected = false
if(arrList.size > 1){
for(i in 0 until arrList.size){
for(j in 0 until arrList.size){
if(i != j && arrList.get(i) == arrList.get(j)){
duplicateDetected = true
}
}
}
}
private boolean isDuplicate() {
for (int i = 0; i < arrayList.size(); i++) {
for (int j = i + 1; j < arrayList.size(); j++) {
if (arrayList.get(i).getName().trim().equalsIgnoreCase(arrayList.get(j).getName().trim())) {
return true;
}
}
}
return false;
}
String tempVal = null;
for (int i = 0; i < l.size(); i++) {
tempVal = l.get(i); //take the ith object out of list
while (l.contains(tempVal)) {
l.remove(tempVal); //remove all matching entries
}
l.add(tempVal); //at last add one entry
}
Note: this will have major performance hit though as items are removed from start of the list.
To address this, we have two options. 1) iterate in reverse order and remove elements. 2) Use LinkedList instead of ArrayList. Due to biased questions asked in interviews to remove duplicates from List without using any other collection, above example is the answer. In real world though, if I have to achieve this, I will put elements from List to Set, simple!
/**
* Method to detect presence of duplicates in a generic list.
* Depends on the equals method of the concrete type. make sure to override it as required.
*/
public static <T> boolean hasDuplicates(List<T> list){
int count = list.size();
T t1,t2;
for(int i=0;i<count;i++){
t1 = list.get(i);
for(int j=i+1;j<count;j++){
t2 = list.get(j);
if(t2.equals(t1)){
return true;
}
}
}
return false;
}
An example of a concrete class that has overridden equals() :
public class Reminder{
private long id;
private int hour;
private int minute;
public Reminder(long id, int hour, int minute){
this.id = id;
this.hour = hour;
this.minute = minute;
}
#Override
public boolean equals(Object other){
if(other == null) return false;
if(this.getClass() != other.getClass()) return false;
Reminder otherReminder = (Reminder) other;
if(this.hour != otherReminder.hour) return false;
if(this.minute != otherReminder.minute) return false;
return true;
}
}
ArrayList<String> withDuplicates = new ArrayList<>();
withDuplicates.add("1");
withDuplicates.add("2");
withDuplicates.add("1");
withDuplicates.add("3");
HashSet<String> set = new HashSet<>(withDuplicates);
ArrayList<String> withoutDupicates = new ArrayList<>(set);
ArrayList<String> duplicates = new ArrayList<String>();
Iterator<String> dupIter = withDuplicates.iterator();
while(dupIter.hasNext())
{
String dupWord = dupIter.next();
if(withDuplicates.contains(dupWord))
{
duplicates.add(dupWord);
}else{
withoutDupicates.add(dupWord);
}
}
System.out.println(duplicates);
System.out.println(withoutDupicates);
A simple solution for learners.
//Method to find the duplicates.
public static List<Integer> findDublicate(List<Integer> numList){
List<Integer> dupLst = new ArrayList<Integer>();
//Compare one number against all the other number except the self.
for(int i =0;i<numList.size();i++) {
for(int j=0 ; j<numList.size();j++) {
if(i!=j && numList.get(i)==numList.get(j)) {
boolean isNumExist = false;
//The below for loop is used for avoid the duplicate again in the result list
for(Integer aNum: dupLst) {
if(aNum==numList.get(i)) {
isNumExist = true;
break;
}
}
if(!isNumExist) {
dupLst.add(numList.get(i));
}
}
}
}
return dupLst;
}