Compare every item to every other item in ArrayList

Question

I'm having trouble with what I thought should be a pretty simple problem.

I need to compare every item in an arrayList with every other item in the the list without comparing items to themselves. It's not as simple as calling an equals() comparison, it involves some custom logic that I've omitted from my code below. Also the ArrayList should not be changed in any way.

The problem I seem to be having is that once I get into the second loop, I don't know if I have another object to compare to (since its a variable sized list).

for(int i =0; i< list.size(); i++){ 
    //get first object to compare to
    String a = list.get(i).getA();

    Iterator itr = list.listIterator(i + 1 ); // I don't know if i + 1 is valid
    while(itr.hasNext()){
        // compare A to all remaining items on list
    }
}

I think I probably going about this the wrong way, I'm open to suggestions or tips on how to do this better.

You say that the ArrayList should not be changed in any way, so the size of the list is not variable : it's constant. And i+1 will thus be a valid index for listIterator(), since i is guaranteed to be < list.size() and listIterator accepts an index up to list.size() inclusive. Your code (once trivial syntax errors are fixed), should thus run as is. — JB Nizet, Feb 14 '11 at 15:45
Over at [CR](https://codereview.stackexchange.com/), this might fit the _Not As Easy As It Looks_ category (especially with a non-symmetric `compare` and general `java.util.List<>`. Starting with `compare every item in [a List L] with every other item in [L]` — greybeard, Jan 11 '16 at 05:47

score 49 · Accepted Answer · answered Feb 14 '11 at 15:38

49

for (int i = 0; i < list.size(); i++) {
  for (int j = i+1; j < list.size(); j++) {
    // compare list.get(i) and list.get(j)
  }
}

answered Feb 14 '11 at 15:38

Kaleb Brasee

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6

This will do more comparisons than required, you will be testing if item1 == item3 and item3 == item1 – Mike Feb 14 '11 at 15:41
IMHO you can even strip out the "if" since j is always at least i+1 and thus they are never equal. – Martin Klinke Feb 14 '11 at 15:42
1

Yeah, just realized that it was doing some unnecessary comparisons, updated the code. – Kaleb Brasee Feb 14 '11 at 15:43
3

@Mike no, because the inner loop only takes elements that are "behind" the current index in the outer loop. – Martin Klinke Feb 14 '11 at 15:43
2

I had it initializing j = 0 at first, then fixed it. – Kaleb Brasee Feb 14 '11 at 15:47
1

Yeah, this actually works, I'm not sure why I was trying to use an iterator, I thought that was the only way to get a sub list. The solution was simple, I was just over thinking this. Thanks – JavaKungFu Feb 14 '11 at 15:51
1

Also, i only needs to go up to list.size()-2. The last element will have already been compared to all other elements during the loop. – chm Sep 05 '12 at 09:45
What's about the complexity of this method? O(nLog(n))? – Zangdak Apr 28 '14 at 13:48
@chm: I thought it was size()-1 (for i). That's at least how I learned bubble sort back in the days... – Erk Jun 24 '18 at 01:42
Hi @KalebBrasee, wondering if same solution can be done using for each in Java 8? – Mahender Reddy Yasa Jan 17 '20 at 12:58

score 3 · Answer 2 · answered Feb 14 '11 at 15:38

3

What's the problem with using for loop inside, just like outside?

for (int j = i + 1; j < list.size(); ++j) {
    ...
}

In general, since Java 5, I used iterators only once or twice.

answered Feb 14 '11 at 15:38

Nikita Rybak

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You should read Effective Java item 46, it explicitly recommends against traditional for loops (whenever possible) – Sean Patrick Floyd Feb 14 '11 at 16:08
@Sean Patrick Floyd, JVM can use intrinsic for that, if need be. I dont say at the moment it does use but the iterator is the easiest thing to be optimized away (via escape analysis) – bestsss Feb 14 '11 at 17:03
2

@bestsss I'm not talking about effective byte code (and neither is Josh Bloch, in this case), but about readable and non-error-prone source code – Sean Patrick Floyd Feb 14 '11 at 17:07

score 2 · Answer 3 · edited Aug 17 '19 at 14:56

2

In some cases this is the best way because your code may have change something and j=i+1 won't check that.

for (int i = 0; i < list.size(); i++){  
    for (int j = 0; j < list.size(); j++) {
                if(i == j) {
               //to do code here
                    continue;
                }

}

}

edited Aug 17 '19 at 14:56

Tariq Mahmood

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answered Jan 10 '16 at 23:22

Piotr Jankowski

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_One_ interpretation of `compare every item in [a List L] with every other item in [L]`. It might be about time to ponder/present a `java.util.stream`-based approach - pity Stream is obliged to be cloneable and provide a useful `equals()` just as much as Iterator. – greybeard Jan 11 '16 at 06:04

BrnLng · Answer 4 · 2016-11-24T20:28:20.810

This code helped me get this behaviour: With a list a,b,c, I should get compared ab, ac and bc, but any other pair would be excess / not needed.

import java.util.*;
import static java.lang.System.out;

// rl = rawList; lr = listReversed
ArrayList<String> rl = new ArrayList<String>();
ArrayList<String> lr = new ArrayList<String>();
rl.add("a");
rl.add("b");
rl.add("c");
rl.add("d");
rl.add("e");
rl.add("f");

lr.addAll(rl);
Collections.reverse(lr);

for (String itemA : rl) {
    lr.remove(lr.size()-1);
        for (String itemZ : lr) {
        System.out.println(itemA + itemZ);
    }
}

The loop goes as like in this picture: Triangular comparison visual example

or as this:

   |   f    e    d    c    b   a
   ------------------------------
a  |  af   ae   ad   ac   ab   ·
b  |  bf   be   bd   bc   ·   
c  |  cf   ce   cd   ·      
d  |  df   de   ·         
e  |  ef   ·            
f  |  ·

total comparisons is a triangular number (n * n-1)/2

score 0 · Answer 5 · answered Jan 06 '16 at 10:38

The following code will compare each item with other list of items using contains() method.Length of for loop must be bigger size() of bigger list then only it will compare all the values of both list.

List<String> str = new ArrayList<String>();
str.add("first");
str.add("second");
str.add("third");
List<String> str1 = new ArrayList<String>();
str1.add("first");
str1.add("second");
str1.add("third1");
for (int i = 0; i<str1.size(); i++)
{
System.out.println(str.contains(str1.get(i)));
}

Output is true true false

score 0 · Answer 6 · edited Jul 22 '21 at 21:47

0

You can do it like this.

for (int i = 0; i < arrayList.size(); i++) {
    for (int j = 0; j < arrayList.size(); j++) {
        if (i!=j &&  arrayList.get(i).YourObjectItem().equals(arrayList.get(j).YourObjectItem())) {
            //Your code will be here
        }

    }
}

edited Jul 22 '21 at 21:47

Reza Rahemtola

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answered Jul 17 '21 at 07:43

Shah Nizar Baloch

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score 0 · Answer 7 · answered Oct 17 '22 at 08:39

Old question, but I just wanted to provide a functional version from the Kaleb Brasee's one, (which, BTF is far more faster and heap friendly than this) for functional programming purists.

    IntStream.range(0, list.size())
        .forEach(i -> compare(list.get(i),
                               list.subList(i + 1, list.size())));
...

private void compare(final Element element, final List<Element> list) {

    list.stream()
        //whatever 

}

just pass it to a set – jmisael56yahoocom Mar 13 '23 at 17:56 — jmisael56yahoocom, Mar 13 '23 at 17:56

score 0 · Answer 8 · answered Mar 13 '23 at 17:56

0

Arraylist<String> list =new Arraylist<>();
list.add("water");
list.add("air");
list.add("earth");
list.add("water");

Set<String> set = new LinkedHashSet<>();
      set.addAll(list);
    System.out.println(set.toString())

answered Mar 13 '23 at 17:56

jmisael56yahoocom

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Compare every item to every other item in ArrayList

8 Answers8

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