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I see a lot of posts with the same topic (mostly with Strings) but haven't found the answer to my question. How would I remove duplicate integers from an ArrayList?

import java.util.*;

public class ArrayList2 {
    public static ArrayList<Integer> removeAllDuplicates(ArrayList<Integer> list) {
        Collections.sort(list);
        for (int i = 0; i < list.size(); i++) {
            if (list.get(i) == list.get(i + 1)) {
                list.remove(i);
            }
        }
    return list;
    }
}

This is the start of my code, the only problem that has arised is that if there are 3 integers with the same value, it only removes one of them. If I put in 4, it removes two of them. PLEASE NO HASHING!!!

The ArrayList and the output when I run it:

List: [-13, -13, -6, -3, 0, 1, 1, 1, 5, 7, 9]
Duplicates Removed: [-13, -6, -3, 0, 1, 1, 5, 7, 9]

This is my first time using this website, so please let me know if I'm doing something wrong with formatting/if there's already an answer to my question that I missed.

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  • 7
    Any specific reason for using a List over a Set? Commented Nov 8, 2018 at 23:20
  • 5
    PLEASE NO HASHING!!! Why? Commented Nov 8, 2018 at 23:22
  • 2
    Would it be more efficient to use a set in this situation? Also, @Micha Wiedenmann, I'll be sure to do that. Commented Nov 8, 2018 at 23:22
  • 1
    @Excel A Set guarantees uniqueness, so yeah, probably - but I'd avoid been tempted to micro-optimise the solution at this point. It would be simpler, easier and more likely to work Commented Nov 8, 2018 at 23:23
  • 1
    new HashSet<>(list) or list.stream().distinct().collect(Collectors.toList()) Commented Nov 8, 2018 at 23:24

4 Answers 4

Reset to default
3

The specific reason why your removeAllDuplicates function doesn't work is that you are still iterating after a successful comparison. If you iterate only when list.get(i) != list.get(i + 1), you will get rid of all the duplicates.

public static ArrayList<Integer> removeAllDuplicates(ArrayList<Integer> list) {
    Collections.sort(list);
    int i = 0;
    while(i < list.size() - 1) {
        if (list.get(i) == list.get(i + 1)) {
            list.remove(i);
        } else {
            i++;
        }
    }
    return list;
}

It's worth noting that the above function is not as fast as it could be. Though the iteration runs quickly enough, the most significant step will be the sort operation (O(n log n)).

To avoid this extra time complexity, consider using a HashSet instead of an ArrayList (if it still fits within the constraints of your problem).

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1 Comment

Checks out! Thank you, I've been stuck on this for a quick minute.
2

Other people have "answered" the basic issue, of the if statement skipping over elements because the for-loop is incrementing the index position, while the size of the array is shrinking.

This is just "another" possible solution. Personally, I don't like mutating Lists in loops and prefer to use iterators, something like...

Collections.sort(list);
Iterator<Integer> it = list.iterator();
Integer last = null;
while (it.hasNext()) {
  Integer next = it.next();
  if (next == last) {
    it.remove();
  }
  last = next;
}

Still, I think some kind of Set would be a simpler and easier solution (and since you'd not have to sort the list, more efficient ;))

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4 Comments

Whether you use an iterator or not, are you not still mutating the array within a loop?
@mettleap Yes, sorry, didn't put that in :P
@unmrshl Not sure what you mean. The iterator is a proxy over the List, it just helps remove the possible "mutation" issues around manipulating Lists within arrays ... or am I thinking of some other constraint. In any case, no forgetting to update the index value or weird manipulation logic for the index. It's just "another" way to achieve the result, which for me, is a little cleaner. (ps - I'm think of for-each loops :P)
You're welcome :) ... without a data structure like a Set, this is pretty efficient
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This assumes that you cannot use a set ("Please No hashing!!!") for a reason such as homework.

Look what happens when you remove a duplicate. Let's say that the dupe you're removing is the 1st of 3 consecutive equal numbers.

                v
-13, -6, -3, 0, 1, 1, 1, 5, 7, 9

Here i is 4 to refer to the first of the 3 1 values. When you remove it, all subsequent elements get shifted down, so that the second 1 value now takes the place of the first 1 value at index i = 4. Then the current iteration ends, another one begins, and i is now 5.

                   v
-13, -6, -3, 0, 1, 1, 5, 7, 9

But now i is referring to the second of the two 1 values left, and the next element 5 isn't equal, so no dupe is found.

Every time you find a duplicate, you must decrease i by 1 so that you can stay on the first element that is a duplicate, to catch 3 or more duplicates in a row.

                v
-13, -6, -3, 0, 1, 1, 5, 7, 9

Now consecutive elements still match, and another 1 value will get removed.

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3 Comments

Would a simple "i--;" in the if statement fix the code? I'm assuming this is what you're saying.
@Excel You "really" want to avoid messing with the iterator of a for-loop
@Excel: It would, but there'd be a bit more work to do...it's also a faux-pas to mess with the iterator of a loop as MadProgrammer alluded to.
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You'd want to have a double for loop as you dont want to check the only the next index but all indexes. As well as remember when you remove a value you want to check that removed value again as it could of been replaced with another duplicate value.

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1 Comment

This would be true if he were not sorting the array. Since it is sorted, the only potential duplicates will be at an index greater than i, which can be handled by simply not iterating after each removal

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