How to remove all the occurences of a element if its duplicate in an arraylist [duplicate]

Question

I have a list with values {"16","b","c","d","e","16","f","g","16","b"}; In this 16 and b are repeated so i want to delete all the entries of them and i need output as c, d, e, f, g. Below program works fine. Any better solutions?

public class Test {

 public static void main(String[] args) {

  ArrayList < String > l = new ArrayList < String > ();
  String[] str = {
   "16",
   "b",
   "c",
   "d",
   "e",
   "16",
   "f",
   "g",
   "16",
   "b"
  };


  for (String s: str) {
   l.add(s);
  }
  List ll = removeDups(l);
  l.removeAll(ll);
  System.out.println("Final List " + l);
 }

 private static List < String > removeDups(ArrayList < String > l) {
  List < String > ll = new ArrayList < String > ();
  for (String a: l) {
   int x = Collections.frequency(l, a);
   if (x > 1) {
    ll.add(a);
   }
  }
  return ll;
 }
}

Kindly explain the code or comment it before posting a question on Stackoverflow. It helps everyone. — Abhishek Agarwal
– Abhishek Agarwal, Commented Jun 15, 2017 at 3:39
As it is, the method removeDups should really be called findDups because it doesn't actually remove anything; it just finds the items that are duplicated. If the line l.removeAll(ll) were moved inside of removeDups, then removeDups would actually merit that name. — Kevin Anderson
– Kevin Anderson, Commented Jun 15, 2017 at 3:57
Considering Collections.frequency is of the order O(n), You can do better by using Hash maps. — Abhishek Agarwal
– Abhishek Agarwal, Commented Jun 15, 2017 at 4:09
I'm voting to close this question as off-topic because it belongs on codereview.stackexchange.com — Dawood ibn Kareem
– Dawood ibn Kareem, Commented Jun 15, 2017 at 4:48
@DawoodibnKareem: This question is not duplicate of stackoverflow.com/questions/203984/…. Neither is the solution. Solution suggested in the link mentioned by you will keep one entry for each repetitive element. But here the issue is to remove every element if it's repeating. It should NOT appear even once in the list. — Keyur Panchal
– Keyur Panchal, Commented Jun 16, 2017 at 8:37

fps · Accepted Answer · 2017-06-15 04:42:43Z

2

One way would be to use streams to find the frequency of each element:

Map<String, Long> counts = yourList.stream()
    .collect(Collectors.groupingBy(
        Function.identity(),     // keep the element as the key
        Collectors.counting())); // values will be the count

Then, you could use removeIf to remove elements based on a condition, for which you'll use the frequencies map calculated above:

yourList.removeIf(elem -> counts.get(elem) > 1);

System.out.println(yourList); // [c, d, e, f, g]

Another way would be to first find out which values have duplicates and which ones are unique. For this, we can use a Map<String, Boolean>:

Map<String, Boolean> duplicates = new LinkedHashMap<>();
yourList.forEach(elem -> duplicates.compute(elem, (k, v) -> v != null));

Here I'm iterating the list and, for each element, I'm putting it into the map, computing the value as true if the element is already present as a key, or false if it's unique.

Then, you could use removeIf on the list, with a predicate that simply returns the value from the map:

yourList.removeIf(duplicates::get);

System.out.println(yourList); // [c, d, e, f, g]

edited Jun 15, 2017 at 4:42

answered Jun 15, 2017 at 4:19

fps

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2 Comments

Sammeta naga srinivas Over a year ago

I think you mean below rite. This is also working fine.for(String s:l){ duplicates.put(s,(duplicates.containsKey(s)?true:false));} Iterator it= l.listIterator(); while(it.hasNext()){ String ss=(String)it.next(); if(duplicates.get(ss)){ it.remove(); } }

fps Over a year ago

@Sammetanagasrinivas Yes, that seems to work too. You don't need the ternary operator. Just duplicates.put(s, duplicate.containsKey(s)); is fine.

Mohit Tyagi · Accepted Answer · 2017-06-19 06:42:15Z

2

You can use Set to remove the duplicate elements from given array list.

Here is the sample code:

Set<String> myStrSet = new HashSet<String>();
Set<String> duplicateSet = new HashSet<String>();

         for(String str : myArrayList){
             if(myStrSet.contains(str)){
                  duplicateSet.add(str);
             } else {
                  myStrSet.add(str);
             }
         }

         for(String str : duplicateSet){
             myStrSet.remove(str);
         }

         for(String str : myStrSet){
             System.out.println("Print non-duplicate elements : " + str);
         }

edited Jun 19, 2017 at 6:42

answered Jun 15, 2017 at 3:49

Mohit Tyagi

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4 Comments

Kevin Anderson Over a year ago

I think OP wanted to remove all instances of array elements which are duplicated; this leaves an instance of each duplicated item.

Sammeta naga srinivas Over a year ago

Yes, I wanted to remove all occurences , Do not want to have any of the occurance if its repeated.

Fady Saad Over a year ago

I don't know how this answer take up vote, It has more than one typo (myStrList and st) and It didn't remove duplicates.

Mohit Tyagi Over a year ago

@Sammetanagasrinivas I have added new code, here first for loop will add duplicate elements into duplicateSet, now in second for loop we will remove the elements from myStrSet with the help of duplicateSet, so finally we will get a list which is free from all the occurrences of duplicate elements.

Keyur Panchal · Accepted Answer · 2017-06-15 07:50:19Z

1

You can compare index and lastIndex for each element. If they are same, the element is unique. We can filter those elements.

// imports
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

// sample code
String[] str = {"16","b","c","d","e","16","f","g","16","b"};
List<String> list = Arrays.asList(str); // List from the array
List<String> newList = new ArrayList<String>();
for(String myStr : list){
    if(list.indexOf(myStr) == list.lastIndexOf(myStr)){
        /*
         * This is a unique element as its index and lastIndex in list are same.
         * Add it to new list.
         */
        newList.add(myStr);
    }
}
// Freeing resources
str = null;
list = null;

System.out.println("Final List: "+ newList);

answered Jun 15, 2017 at 7:50

Keyur Panchal

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2 Comments

Sammeta naga srinivas Over a year ago

Wow. its working fine. Thank you so much.

Keyur Panchal Over a year ago

@Sammetanagasrinivas Can you please upvote the answer and accept the answer as solution ?

Arun Sudhakaran · Accepted Answer · 2017-06-15 05:19:15Z

0

I think this will do

public class DeleteDuplicates {

    public static void main(String[] args) {

        String[] str={"16","b","c","d","e","16","f","g","16","b"};
        List<String> l= new ArrayList<String>();
        Set<String> set = new HashSet<String>();

        for(String string : str) {

            if(set.add(string))
                l.add(string);
            else
                l.remove(string);
        }              

        System.out.println(l);
    }
}

answered Jun 15, 2017 at 5:19

Arun Sudhakaran

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