Arrays Equals Ignoring Order

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Java: Checking equality of arrays (order doesnt matter)

I have two arrays :

String[] a1 = {"a", "b", "c"};
String[] a2 = {"c", "b", "a"};

I need to check if both contains same elements (and of same length) irrespective of order of elements.

I tried Arrays.equals(a1, a2) but it considers order of element. org.apache.commons.lang.ArrayUtils does not provide this thing.

I know I can achieve the same by creating my own method (checking for same length, then sorting both array and then using Arrays.equals(a1, a2)) but wanted to know if this thing is provided in any API or there is more smart way to do the same.

Have you tried Arrays.sort(a2), and then check to see if they are equal? — CBredlow, Aug 17 '12 at 16:16
You could sort the arrays before you call Arrays.equals but this would be costly. — Denys Séguret, Aug 17 '12 at 16:16
A probably more efficient solution would be to add the items of first array to a hashset and then check what items of the second arrays are yet in the hashset. — Denys Séguret, Aug 17 '12 at 16:18

BlackVegetable · Accepted Answer · 2016-09-03T13:27:06.720

35

If you have these arrays in something inheriting from Collection, you can just use collection.containsAll( otherCollection ) from the Collection interface. However, you'll also need to compare the lengths of the two to verify that one set isn't a superset of the other.

(Thanks go to Aardvarkk and piegames.)

http://docs.oracle.com/javase/6/docs/api/java/util/Collection.html#containsAll(java.util.Collection)

Note: This will work up to a point. This is defined to check for at least one of any element existing. That is, if you have 3 a values in one collection, and 7 a values in the other, that will not necessarily cause it to call them unequal.

Examples:

[a, b, c] == [c, a, b]             // Works -- Mixed order
[a, b, c, d, d] == [a, b, d, c, d] // Works -- Mixed order with repeats
[a, b, c, d, d] == [a, b, b, c, d] // FAILS -- Different repeats
[a, b, c, d, d] != [a, b, c, d]    // Works -- Length differs with repeats
[a, b, c, d] != [a, b, c]          // Works -- Length differs
[a, b, d] != [a, b, c]             // Works -- Disjoint sets

edited Sep 03 '16 at 13:27

answered Aug 17 '12 at 16:19

BlackVegetable

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Wouldn't you need to check containsAll in both directions? A one-way evaluation would return true for a set which was a superset of the other, which doesn't appear to be what the OP wants. – aardvarkk Aug 17 '12 at 16:21
@aardvarkk Excellent point! I'll edit my post. – BlackVegetable Aug 17 '12 at 16:26
4

This doesn't handle duplicates correctly. The JavaDoc doesn't make this clear, but `a.containsAll(b)` iff `a` contains **at least one** `x` for each element `x` in `b`. As an example, try testing with `a={5}`, `b={5,5}`. They're not equal when viewed as multisets, but `containsAll` returns `true` in both directions. – Daniel Lubarov Nov 14 '13 at 01:31
At least this is the interpretation of `containsAll` that's used in practice. Even Guava's [Multiset.containsAll](http://google-collections.googlecode.com/svn/trunk/javadoc/com/google/common/collect/Multiset.html#containsAll) "does not take into account the occurrence count of an element". – Daniel Lubarov Nov 14 '13 at 01:35
@Daniel Right, this was covered by the portion of my answer labeled "Note". – BlackVegetable Nov 14 '13 at 02:26
2

@aardvarkk Wouldn't it be enough to just check the length of both Arrays instead of the second containsAll? (Assuming there are no duplicates) – piegames Sep 03 '16 at 09:17
@piegames That's absolutely correct. I'm a little embarrassed I didn't think of that. I will update this answer. – BlackVegetable Sep 03 '16 at 13:20

score 28 · Answer 2 · answered Aug 18 '12 at 05:45

28

i think it may work for you, sort first array with

Arrays.sort(Object[]);

and after that you can compare with

Arrays.equals(Object[],Object[]);

complete code is:

String[] a1 = {"a", "b", "c"};
String[] a2 = {"c", "b", "a"};
Arrays.sort(a2);

boolean result= Arrays.equals(a1, a2);

answered Aug 18 '12 at 05:45

satvinder singh

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Sorting should happen on both a1 and a2. There is no guarantee a1 is already sorted. – Gaʀʀʏ May 27 '16 at 15:22
Even then, sorting isn't always a good idea: it mutates the data. If you don't want it you have to duplicate the data which costs time and memory. – piegames Sep 04 '16 at 16:24
This won't work if the objects don't implement comparable – Dylanthepiguy Apr 12 '17 at 08:05

score 15 · Answer 3 · edited Aug 17 '12 at 16:44

15

Convert the lists to sets before comparing them :

new HashSet( Arrays.asList( a1 )).equals( new HashSet( Arrays.asList( a2 ) ));

Alternatively, you can sort the arrays using Arrays.sort() but that might break code which depends on the order of the elements in the arrays.

edited Aug 17 '12 at 16:44

Wandang

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answered Aug 17 '12 at 16:18

Aaron Digulla

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That only works if there are no duplicates. – assylias Aug 17 '12 at 16:19
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This is also more efficient as it runs O(m+n) whereas doing containsAll on a list is O(n^2+m^2) – Martin Andersson Nov 18 '13 at 11:07

score 2 · Answer 4 · answered Aug 17 '12 at 16:23

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Use java.util.Set method equals. Compares two sets have the same size, and every member of the specified set is contained in other set.

answered Aug 17 '12 at 16:23

Roman C

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Arrays Equals Ignoring Order

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