Java: Checking equality of arrays (order doesn't matter)

• Arrays.sort(s1);
• Arrays.sort(s2);
• Arrays.equals(s1,s2);

In case you do not want to modify the original arrays

 Arrays.equals( Arrays.sort( Arrays.copyof(s1,s1.length)),
                Arrays.sort( Arrays.copyof(s2,s2.length)) );

Arrays.sort() uses an optimized quick sort which is nlog(n) for average but O(n2) in worst case. From the java docs. So the worst case it will O(n2) but practically it will be O(nlogn) for most of the cases.

The sorting algorithm is a tuned quicksort, adapted from Jon L. Bentley and M. Douglas McIlroy's "Engineering a Sort Function", Software-Practice and Experience, Vol. 23(11) P. 1249-1265 (November 1993). This algorithm offers n*log(n) performance on many data sets that cause other quicksorts to degrade to quadratic performance.

Others have suggested sorting the arrays. But since you're looking for the "cleanest" solution, I think the original arrays shouldn't be touched. Hence:

List<String> l1 = new ArrayList<String>(Arrays.asList(s1));
List<String> l2 = new ArrayList<String>(Arrays.asList(s2));

Collections.sort(l1);
Collections.sort(l2);

boolean outcome = l1.equals(l2);

If you are using Eclipse Collections, you can use a Bag to figure out if the two arrays are equal.

String[] s1 = {"a", "b", "c", "c"};
String[] s2 = {"c", "a", "b", "c"};

Bag<String> h1 = Bags.mutable.with(s1);
Bag<String> h2 = Bags.mutable.with(s2);
Assert.assertEquals(h1, h2);

Bags (also known as multisets) are considered equal if they have the same number of occurrences of each element. Order doesn't matter, and it properly handles duplicate elements. The advantage of using a bag backed by a hashtable is that creating one takes linear time. Sorting both takes O(n log n).

Note: I am a committer for Eclipse Collections